**CBSE NCERT NOTES CLASS 12 PHYSICS CHAPTER 4**

**MOVING CHARGES AND MAGNETRISM**

**Magnetic Field**

The space in the surrounding of a magnet or any current carrying conductor in which its magnetic influence can be experienced.

**Oersted’s Observation**

A **magnetic field** is produced in the surrounding of any current carrying conductor. The direction of this magnetic field can be obtained by Ampere’s swimming rule.

SI unit of magnetic field is Wm^{-2} or T (telsa).

The strength of magnetic field is one Tesla, if a charge of one coulomb, when moving with a velocity of 1 ms^{-1} along a direction perpendicular to the direction of the magnetic field experiences a force of one Newton.

1 Tesla (T) = 1 Weber metre^{-2}(Wm^{-2})

= 1 Newton Ampere^{-1} metre^{-1 }(NA^{-1}m^{-1})

CGS unit of magnetic field is called Gauss or Oersted.

1 Gauss = Oersted = 10^{-4} Tesla.

**Ampere’s Swimming Rule (For direction of deflection of compass needle)**

If a man is swimming along the wire in the direction of current his face turned towards the needle, so that the current enters through his feet, then northpole of the magnetic needle will be deflected towards his left hand.

The conventional sign for a magnetic field coming out of the plane and normal to it is a **dot i.e., ‘.’**, the magnetic field perpendicular to the plane in the downward direction is denoted by **cross, i.e., ‘×’**.

**Magnetic field due to a current element - Biot Savart’s law**

The magnetic field produced by a current carrying element of length *dl*, carrying current *I* at a point separated by a distance *r* is given by

$$\mathrm{d}\overrightarrow{\mathrm{B}}\mathrm{}=\mathrm{}\mathrm{}\mathrm{}\frac{{\mathrm{\mu}}_{\mathrm{o}}\mathrm{I}}{4\mathrm{\pi}}\frac{\overrightarrow{\mathrm{d}\mathrm{l}}\times \overrightarrow{\mathrm{r}}}{{\mathrm{r}}^{3}}$$

Or

$$\mathrm{d}\mathrm{B}\mathrm{}=\mathrm{}\mathrm{}\mathrm{}\frac{{\mathrm{\mu}}_{\mathrm{o}}}{4\mathrm{\pi}}\frac{\mathrm{I}\mathrm{}\mathrm{d}\mathrm{l}\mathrm{}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{\theta}}{{\mathrm{r}}^{2}}\mathrm{}\mathrm{}$$

where,

θ is the angle between the direction of the current and $\overrightarrow{\mathrm{r}}$

μo is **absolute permeability** of the free space.

The direction of magnetic field $\mathrm{d}\overrightarrow{\mathrm{B}}\mathrm{}$is that of *d*$\overrightarrow{\mathrm{l}}\times \overrightarrow{\mathrm{r}}$.

Value of

$$\mathrm{}{\frac{{\mathrm{\mu}}_{\mathrm{o}}}{4\mathrm{\pi}}=10}^{-7}\mathrm{}\mathrm{}\mathrm{T}\mathrm{m}/\mathrm{A}$$

$${\mathrm{\varepsilon}}_{\mathrm{o}}{\mathrm{\mu}}_{\mathrm{o}}=4\mathrm{\pi}{\mathrm{\varepsilon}}_{\mathrm{o}}\mathrm{}\left(\frac{{\mathrm{\mu}}_{\mathrm{o}}}{4\mathrm{\pi}}\right)=\mathrm{}\frac{1}{9\times {10}^{9}}\times {10}^{-7}\mathrm{}{\mathrm{m}}^{-2}$$

$$=\frac{1}{{\left(3\times {10}^{8}\right)}^{2}}=\frac{1}{{\mathrm{c}}^{2}}\mathrm{}{\mathrm{m}}^{-2}$$

**Comparision of Electrostatic and Magnetic Force**

- Both are long range, since both depend inversely on the square of distance from the source to the point of interest.
- The principle of superposition applies to both fields.
- The electrostatic field is produced by a scalar source, namely, the electric charge. The magnetic field is produced by a vector source IdL.
- The electrostatic field is along the displacement vector joining the source and the point. The magnetic field is perpendicular to the plane containing the displacement vector r and the current element IdL.
- There is an angle dependence in the Biot-Savart law which is not present in the electrostatic case.

**Magnetic Field Lines**

- Magnetic field lines are imaginary closed curves around magnet. They are used to represent magnetic field B in a region.
- Tangent drawn at any point to the curve gives the direction of magnetic field.
- They do not intersect each other (why ?).
- Outside a magnet, they are directed from north to south pole and inside a magnet they are directed from south to north pole.
- Density of magnetic field lines is higher near the magnet and lower as we move away.

**Magnetic Field Due to a Straight Current Carrying Conductor**

The magnetic field lines due to a straight current carrying conductor are concentric circles having centre on the conductor and in a plane perpendicular to the conductor.

**Maxwell’s Cork Screw Rule **

If a right handed cork screw is imagined to be rotated in such a direction that tip of the screw points in the direction of the current, then direction of rotation of thumb gives the direction of magnetic field lines.

**Right hand thumb rule for straight current carrying wire**

To determine the direction of the magnetic field due to a long wire, the **right hand thumb rule** is,

Grasp the wire in your right hand with your extended thumb pointing in the direction of the current. Your fingers will curl around in the direction of the magnetic field.

**Strength of magnetic field due to straight current carrying conductor**

$$\mathrm{B}\mathrm{}=\frac{\mathrm{\mu}\mathrm{o}}{4\mathrm{\pi}}\mathrm{}\frac{\mathrm{I}}{\mathrm{r}}(\mathit{sin}{\mathrm{\varphi}}_{1}+\mathit{sin}{\mathrm{\varphi}}_{2})$$

where ϕ_{1} and ϕ_{2} are angles, which the lines joining the two ends of the conductor to the observation point make with the perpendicular from the observation point to the conductor.

**Proof**

Let us consider a straight conductor carrying current I. We need to find the magnetic field at a point P, which is at a perpendicular distance r from the conductor. Point P makes angles ϕ_{1} and ϕ_{2} with the perpendicular to the conductor.

Now let us consider a current element on the conductor. The line joining it with P makes an angle θ with the perpendicular.

From the geometry of the figure, we can see that,

$$\mathrm{l}\mathrm{}=\mathrm{}\mathrm{r}\mathrm{tan\; \theta}\mathrm{}$$

Differentiating this equation we get,

$$\mathrm{d}\mathrm{l}\mathrm{}=\mathrm{}\mathrm{r}{\mathrm{sec}}^{2}\theta \mathrm{}\mathrm{d\theta}$$

$$\mathrm{P}\mathrm{Q}\mathrm{}=\mathrm{}\mathrm{r}\mathrm{cos\; \theta}\mathrm{}$$

Now, as per Biot Savart’s law, the field at P due to this element, will be

$$\mathrm{d}\mathrm{B}\mathrm{}=\mathrm{}\mathrm{}\mathrm{}\frac{{\mathrm{\mu}}_{\mathrm{o}}}{4\mathrm{\pi}}\frac{\mathrm{I}\mathrm{}\mathrm{d}\mathrm{l}\mathrm{}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{}\left(\frac{\mathrm{\pi}}{2}-\mathrm{\theta}\right)}{{\mathrm{P}\mathrm{Q}}^{2}}=\mathrm{}\mathrm{}\mathrm{}\frac{{\mathrm{\mu}}_{\mathrm{o}}}{4\mathrm{\pi}}\frac{\mathrm{I}\mathrm{}\mathrm{d}\mathrm{l}\mathrm{cos\; \theta}\mathrm{}\mathrm{}}{{\mathrm{P}\mathrm{Q}}^{2}}$$

Putting the value of *dl* and *PQ, *we get,

$$\mathrm{d}\mathrm{B}\mathrm{}=\mathrm{}\mathrm{}\mathrm{}\frac{{\mathrm{\mu}}_{\mathrm{o}}}{4\mathrm{\pi}}\frac{\mathrm{I}\mathrm{}\mathrm{r}{\mathrm{sec}}^{2}\mathrm{\theta}\mathrm{}\mathrm{}\mathrm{d\theta}\mathrm{}\times \mathrm{cos\; \theta}}{{\mathrm{r}}^{2}{\mathrm{cos\; \theta}}^{2}}$$

$$=\mathrm{}\mathrm{}\mathrm{}\frac{{\mathrm{\mu}}_{\mathrm{o}}\mathrm{I}\mathrm{}}{4\mathrm{\pi}\mathrm{r}}\mathrm{cos\; \theta}\mathrm{}\mathrm{d}\mathrm{}$$

Integrating the above equation, we get,

$${\int}_{0}^{\mathrm{B}}\mathrm{d}\mathrm{B}=\mathrm{}\mathrm{}\mathrm{}\frac{{\mathrm{\mu}}_{\mathrm{o}}\mathrm{I}\mathrm{}}{4\mathrm{\pi}\mathrm{r}}{\int}_{{-\mathrm{\varphi}}_{1}}^{{\mathrm{\varphi}}_{2}}\mathrm{cos\; \theta}\mathrm{}\mathrm{d\theta}\mathrm{}$$

$$\mathrm{B}=\mathrm{}\mathrm{}\mathrm{}\frac{{\mathrm{\mu}}_{\mathrm{o}}\mathrm{I}\mathrm{}}{4\mathrm{\pi}\mathrm{r}}{\left[\mathrm{sin}\right]}_{{-\mathrm{\varphi}}_{1}}^{{\mathrm{\varphi}}_{2}}$$

$$\mathrm{B}\mathrm{}=\frac{\mathrm{\mu}\mathrm{o}}{4\mathrm{\pi}}\mathrm{}\frac{\mathrm{I}}{\mathrm{r}}(\mathit{sin}{\mathrm{\varphi}}_{2}-\mathit{sin}(-{\mathrm{\varphi}}_{1}))$$

$$\mathrm{B}\mathrm{}=\frac{\mathrm{\mu}\mathrm{o}}{4\mathrm{\pi}}\mathrm{}\frac{\mathrm{I}}{\mathrm{r}}(\mathit{sin}{\mathrm{\varphi}}_{1}+\mathit{sin}{\mathrm{\varphi}}_{2})$$

- For infinite length conductor and observation point is near the centre of the conductor,
$${\mathrm{\varphi}}_{1}\mathrm{}=\mathrm{}\frac{\mathrm{\pi}}{2}$$

$${\mathrm{\varphi}}_{2}\mathrm{}=\mathrm{}\frac{\mathrm{\pi}}{2}$$

$$\mathrm{B}\mathrm{}=\frac{\mathrm{\mu}\mathrm{o}}{4\mathrm{\pi}}\mathrm{}\frac{\mathrm{I}}{\mathrm{r}}\left(1+1\right)$$

$$\Rightarrow \mathrm{B}=\frac{\mathrm{\mu}\mathrm{o}}{2\mathrm{\pi}}\mathrm{}\frac{\mathrm{I}}{\mathrm{r}}\mathrm{}$$

- For infinite length conductor and observation point is near one end of the conductor,
$${\mathrm{\varphi}}_{1}\mathrm{}=\mathrm{}\frac{\mathrm{\pi}}{2}$$

$${\mathrm{\varphi}}_{2}\mathrm{}=\mathrm{}0$$

$$\mathrm{B}\mathrm{}=\frac{\mathrm{\mu}\mathrm{o}}{4\mathrm{\pi}}\mathrm{}\frac{\mathrm{I}}{\mathrm{r}}$$

**Magnetic field on the axis of a current carrying circular coil**

Consider a point P on the axis of the current carrying coil, at a distance x from the centre of the coil.

From the geometry,

The angle between $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{d}\mathrm{l}}$ is 90°.

Hence field due to current carrying element $\overrightarrow{\mathrm{d}\mathrm{l}}$ is

$$\mathrm{d}\mathrm{B}\mathrm{}=\mathrm{}\mathrm{}\frac{\mathrm{\mu}\mathrm{o}}{4\mathrm{\pi}}\mathrm{}\frac{\mathrm{I}\mathrm{}\mathrm{d}\mathrm{l}\mathrm{}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{}90\xb0}{{\mathrm{r}}^{2}}=\mathrm{}\frac{\mathrm{\mu}\mathrm{o}}{4\mathrm{\pi}}\mathrm{}\frac{\mathrm{I}\mathrm{}\mathrm{d}\mathrm{l}\mathrm{}}{{\mathrm{r}}^{2}}$$

The direction of magnetic field is perpendicular to the plane containing $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{d}\mathrm{l}}$ . So the magnetic field $\overrightarrow{\mathrm{d}\mathrm{B}}$ has two components,

*dB cos θ* along the Y – axis

*dB sin θ* along the X - axis

Similarly, consider another current carrying element $\overrightarrow{\mathrm{d}\mathrm{l}}\mathrm{\text{'}}$, which is diametrically opposite to the point A. The magnetic field due to this current carrying element, $\overrightarrow{\mathrm{d}\mathrm{B}}\mathrm{\text{'}}$ also has two components

*- dB’ cos θ* along the Y – axis

*dB’ sin θ* along the X - axis

Since *dB cos θ* and *dB’ cos* *θ* are equal in magnitude and opposite in direction, they cancel each other.

On the other hand, the components *dB sin θ* and *dB’ sin θ* are equal in magnitude and in same direction so they add up.

Hence total magnetic field due to the circular current carrying coil at the axis is,

$$\mathrm{B}\mathrm{}=\int \mathrm{d}\mathrm{B}\mathit{sin}=\mathrm{}{\int}_{0}^{2\mathrm{\pi}\mathrm{a}}\frac{{\mathrm{\mu}}_{\mathrm{o}}}{4\mathrm{\pi}}\frac{\mathrm{I}\mathrm{}\mathrm{d}\mathrm{l}\mathrm{}}{{\mathrm{r}}^{2}}\frac{\mathrm{a}}{\mathrm{r}}\mathrm{}$$

$$=\mathrm{}{\int}_{0}^{2\mathrm{\pi}\mathrm{a}}\frac{{\mathrm{\mu}}_{\mathrm{o}}}{4\mathrm{\pi}}\frac{\mathrm{I}\mathrm{}\mathrm{d}\mathrm{l}\mathrm{}}{{\mathrm{r}}^{2}}\times \frac{\mathrm{a}}{\mathrm{r}}\mathrm{}\mathrm{}\mathrm{}$$

Now,

$$\mathit{sin}=\frac{\mathrm{a}}{\mathrm{r}}\mathrm{}$$

and

* *$\mathrm{r}\mathrm{}=\mathrm{}\sqrt{{\mathrm{a}}^{2}+{\mathrm{x}}^{2}}$

Therefore,

$$\mathrm{B}\mathrm{}=\mathrm{}{\int}_{0}^{2\mathrm{\pi}\mathrm{a}}\frac{{\mathrm{\mu}}_{\mathrm{o}}}{4\mathrm{\pi}}\frac{\mathrm{I}\mathrm{}\mathrm{a}\mathrm{}\mathrm{d}\mathrm{l}}{{\left({\mathrm{a}}^{2}+{\mathrm{x}}^{2}\right)}^{\frac{3}{2}}}$$

$$=\frac{{\mathrm{\mu}}_{\mathrm{o}}}{4\mathrm{\pi}}\frac{\mathrm{I}\mathrm{}\mathrm{a}\mathrm{}}{{({\mathrm{a}}^{2}+{\mathrm{x}}^{2})}^{\frac{3}{2}}}\mathrm{}{\int}_{0}^{2\mathrm{\pi}\mathrm{a}}\mathrm{d}\mathrm{l}$$

$$=\frac{{\mathrm{\mu}}_{\mathrm{o}}}{4\mathrm{\pi}}\times \frac{\mathrm{I}\mathrm{}\mathrm{a}\mathrm{}}{{\left({\mathrm{a}}^{2}+{\mathrm{x}}^{2}\right)}^{\frac{3}{2}}}\times 2\mathrm{\pi}\mathrm{a}$$

$$\Rightarrow \mathrm{}\mathrm{B}=\frac{{\mathrm{\mu}}_{\mathrm{o}}}{2}\frac{\mathrm{I}\mathrm{}{\mathrm{a}}^{2}\mathrm{}}{{({\mathrm{a}}^{2}+{\mathrm{x}}^{2})}^{\frac{3}{2}}}$$

**Magnetic field at centre of the coil, **

$$\mathrm{x}\mathrm{}=\mathrm{}0\mathrm{}\Rightarrow \mathrm{}\mathrm{B}=\frac{{\mathrm{\mu}}_{\mathrm{o}}}{2}\frac{\mathrm{I}\mathrm{}\mathrm{}}{\mathrm{a}}$$

If there are ‘*n*’ number of turns in the circular coil, then the magnetic field is

$$\mathrm{B}=\frac{{\mathrm{\mu}}_{\mathrm{o}}}{2}\frac{\mathrm{n}\mathrm{}\mathrm{I}\mathrm{}\mathrm{}}{\mathrm{a}}$$

**Right hand thumb rule for current carrying loop**

Curl the palm of your right hand around the circular wire with the fingers pointing in the direction of the current. The right-hand thumb gives the direction of the magnetic field.

**Ampere’s circuital law**

The line integral of magnetic field induction B around any closed path in vacuum is equal to μ_{0} times the total current threading the closed path (or passing through the surface enclosed by the closed path), i.e.,

$$\oint \overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{d}\mathrm{l}}=\overrightarrow{\mathrm{B}}\mathrm{}\oint \overrightarrow{\mathrm{d}\mathrm{l}}\mathrm{}=\frac{\mathrm{}{\mathrm{\mu}}_{\mathrm{o}}\mathrm{I}}{2\mathrm{\pi}\mathrm{r}}\times 2\mathrm{\pi}\mathrm{r}=\mathrm{}{\mathrm{\mu}}_{\mathrm{o}}\mathrm{I}$$

where $\overrightarrow{\mathrm{B}}$ is the magnetic field, $\overrightarrow{\mathrm{d}\mathrm{l}}$ is small length element, *μ*_{o} is the absolute permeability of free space and *I* is the current.

- Ampere’s circuital law holds good for a closed path of any size and shape around a current carrying conductor because the relation is independent of distance form conductor.
- In practice, the Amperian loop is chosen in such a way that one of the following conditions is satisfied.
- B is tangential to the loop and is a non-zero constant B, or
- B
- B

**Magnetic field due to current carrying wire of infinite length using Ampere’s circuital law**

- The field at every point on a circle of radius r, (with the wire along the axis), is same in magnitude. In other words, the magnetic field possesses a
**cylindrical symmetry**.$$\oint \overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{d}\mathrm{l}}=\mathrm{}{\mathrm{\mu}}_{\mathrm{o}}\mathrm{I}\mathrm{}$$

$$\Rightarrow \mathrm{B}\mathrm{}\times \mathrm{}2\mathrm{\pi}\mathrm{r}\mathrm{}=\mathrm{}{\mathrm{\mu}}_{\mathrm{o}}\mathrm{I}\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{B}\mathrm{}=\frac{{\mathrm{\mu}}_{\mathrm{o}}\mathrm{I}}{2\mathrm{\pi}\mathrm{r}}$$

- The field that normally can depend on three coordinates depends only on one:
*r*. - The field direction at any point on this circle is tangential to it. Thus, the lines of constant magnitude of magnetic field form concentric circles.
- Even though the wire is infinite, the field due to it at a nonzero distance is not infinite. It tends to blow up only when we come very close to the wire.
- The field is directly proportional to the current and inversely proportional to the distance from the (infinitely long) current source.
**Ampere’s law**in magnetism is analogous to**Gauss’s law**in electrostatics. In order to apply them, the system must possess certain symmetry.- In the case of an infinite wire, the system possesses cylindrical symmetry and Ampere’s law can be readily applied. However, when the length of the wire is finite, Biot-Savart law must be used instead.
- This law is usually used to find magnetic field only in special cases when the contour integral can be found as a function of single field value based on symmetry.
- Magnetic field of a circular current loop is not so simple and Ampere's law cannot be easily used to find it. In such cases, the method of choice is to use the Biot-Savart law (integrate the contributions to the field due to elements of the circuit – as derived earlier).

**Magnetic field outside a current carrying wire **

Let us consider a cylindrical conductor of radius ‘*a*’, carrying current I.

Take an Amperian loop of radius *r > a.*

The current passing through the circular cross section of radius *r *is *I*

Now from Amperes’ circuit law, we have*,*

$${\oint}_{0}^{2\mathrm{\pi}\mathrm{R}}\overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{d}\mathrm{l}}=\mathrm{}{\mathrm{\mu}}_{\mathrm{o}}\mathrm{I}\mathrm{}$$

$$\Rightarrow \mathrm{B}\mathrm{}\times \mathrm{}2\mathrm{\pi}\mathrm{r}\mathrm{}=\mathrm{}{\mathrm{\mu}}_{\mathrm{o}}\mathrm{I}\mathrm{}$$

$$\Rightarrow \mathrm{}\mathrm{B}\mathrm{}=\frac{{\mathrm{\mu}}_{\mathrm{o}}\mathrm{I}}{2\mathrm{\pi}\mathrm{r}}$$

Therefore, magnetic field outside the wire *B =* $\frac{{\mathrm{\mu}}_{\mathrm{o}}\mathrm{I}}{2\mathrm{\pi}\mathrm{r}}$

**Magnetic field inside a current carrying wire **

Let us consider a cylindrical conductor of radius ‘*a*’, carrying current *I*.

Take an Amperian loop of radius *r < a.*

The current density through the conductor

$$\mathrm{J}=\frac{\mathrm{I}}{\mathrm{\pi}{\mathrm{a}}^{2}}$$

Hence current passing through the circular cross section of radius *r *

$${\mathrm{I}}_{1}=\mathrm{}\mathrm{\pi}{\mathrm{r}}^{2}\times \frac{\mathrm{I}}{\mathrm{\pi}{\mathrm{a}}^{2}}=\frac{\mathrm{I}{\mathrm{r}}^{2}}{{\mathrm{a}}^{2}}$$

Now from Amperes’ circuit law, we have*,*

$${\oint}_{0}^{2\mathrm{\pi}\mathrm{r}}\overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{d}\mathrm{l}}=\mathrm{}{\mathrm{\mu}}_{\mathrm{o}}\mathrm{}{\mathrm{I}}_{1}\mathrm{}$$

$$\Rightarrow \mathrm{B}\mathrm{}\times \mathrm{}2\mathrm{\pi}\mathrm{r}\mathrm{}=\mathrm{}\frac{{\mathrm{\mu}}_{\mathrm{o}}\mathrm{}\mathrm{I}{\mathrm{}\mathrm{r}}^{2}}{{\mathrm{a}}^{2}}$$

$$\Rightarrow \mathrm{}\mathrm{B}\mathrm{}=\frac{{\mathrm{\mu}}_{\mathrm{o}}\mathrm{}\mathrm{I}\mathrm{}{\mathrm{r}}^{2}}{{\mathrm{a}}^{2}}\times \frac{1}{2\mathrm{\pi}\mathrm{r}}$$

Therefore, magnetic field inside the wire *B =* $\frac{{\mathrm{\mu}}_{\mathrm{o}}\mathrm{}\mathrm{I}\mathrm{}\mathrm{r}}{2\mathrm{\pi}{\mathrm{a}}^{2}}$

**SOLENOID**

A solenoid is a closely wound helix of insulated copper wire. It behaves like magnet when current passes through it.

A long solenoid means that its length is much larger compared to its radius.

Each turn can be regarded as a circular loop. The net magnetic field is the vector sum of the fields due to all the turns.

**The magnetic field outside the solenoid is practically zero.**

We have two "rows" of currents, the top one (current coming out of the page) and the bottom one (current going into the page). We follow the "right hand thumb rule", and take into account the fact that the fields from all of the individual loops add together.

Inside the solenoid, the fields from the two rows point in the same direction, reinforcing each other. This results in a net field that is strong.

Outside the solenoid, the fields due to the circular loops between neighbouring turns tend to cancel each other.

The top row produces a combined magnetic field that points (mostly) left above itself, and (mostly) right below itself.

The field due to bottom row points (mostly) right above itself, and (mostly) left below itself.

The fields at any point outside the solenoid, from the two rows, point in **opposite** directions, mostly canceling each other (totally canceling for an infinite solenoid).

Therefore the magnetic field outside a solenoid is weak or almost zero.

**The magnetic field inside the solenoid**

Let us consider a rectangular Amperian loop ** abcd** as shown in the diagram.

The field is zero along ** cd** (outside the solenoid).

The field component is zero along transverse sections ** bc** and

**,. Thus, these two sections make no contribution.**

*ad*Let the field along ** ab** be

*B*. Thus, the relevant length of the Amperian loop is,

*L = h.*

Let n be the number of turns per unit length, then the total number of turns is *nh*. The enclosed current is, *I*_{e}* = I nh*, where I is the current in the solenoid.

From Ampere’s circuital law

*BL = μ*_{o}*I*_{e}*, ⇒ B h = μ*_{o}*I (n h) *

*or B = μ _{o} n I*

Magnetic field at a point on one end of a long solenoid is given by

$$\mathrm{B}\mathrm{}=\frac{\mathrm{\mu}\mathrm{o}\mathrm{}\mathrm{n}\mathrm{}\mathrm{I}}{2}$$

**TOROID**

A toroidal solenoid is an anchor ring around which large number of turns of a copper wire are wrapped. A toroid is an endless solenoid in the form of a ring.

The direction of the magnetic field inside is clockwise as per the right-hand thumb rule for circular loops. Let us consider three circular Amperian loops 1, 2 and 3.

By symmetry, the magnetic field should be tangential to each of them and constant in magnitude for a given loop

Loop 1 encloses no current, hence *B*_{1} = 0

For loop 3, the current coming out of the plane of the paper is cancelled exactly by the current going into it. Hence *B*_{3} at Q is also zero.

For loop 2, the current enclosed *I*_{e} is (for *N* turns of toroidal coil) *N I*, that is

$$\mathrm{B}\mathrm{}\left(2\mathrm{\pi}\mathrm{r}\right)\mathrm{}=\mathrm{}{\mathrm{\mu}}_{\mathrm{o}}\mathrm{N}\mathrm{I}$$

$$\Rightarrow \mathrm{B}\mathrm{}=\frac{{\mathrm{\mu}}_{\mathrm{o}}\mathrm{N}\mathrm{I}}{2\mathrm{\pi}\mathrm{r}}$$

If we take,

$$\mathrm{}\mathrm{n}\mathrm{}=\frac{\mathrm{N}}{2\mathrm{\pi}\mathrm{r}}\Rightarrow \mathrm{}\mathrm{N}\mathrm{}=2\mathrm{\pi}\mathrm{r}\mathrm{n},$$

we get,

* B = μ*_{o}*n I *

Therefore, the field inside the toroid is the same as the field inside the solenoid.

Magnetic field inside a toroid is constant and is always tangential to the circular closed path.

Magnetic field at any point inside the empty space surrounded by the toroid and outside the toroid, is zero, because net current enclosed by this space is zero.

**MAGNETIC FORCE ON A MOVING CHARGE**

A point charge *q* (moving with a velocity *v* and, located at *r* at a given time *t*) in presence of both the electric field *E(r) *and the magnetic field *B(r)*, experiences force due to both electrical and magnetic field. The force on an electric charge q due to both electrical and magnetic field is called the **Lorentz force** and can be written as

$$\overrightarrow{{\mathrm{F}}_{\mathrm{E}}}\mathrm{}=\mathrm{q}\overrightarrow{\mathrm{E}}$$

And

$$\overrightarrow{{\mathrm{F}}_{\mathrm{B}}}\mathrm{}=\mathrm{q}\mathrm{}\overrightarrow{\mathrm{v}}\times \mathrm{}\overrightarrow{\mathrm{B}}\mathrm{}$$

Total force, $\overrightarrow{\mathrm{F}}\mathrm{}=\overrightarrow{{\mathrm{F}}_{\mathrm{E}}}\mathrm{}+\mathrm{}\overrightarrow{{\mathrm{F}}_{\mathrm{B}}}=\mathrm{q}[\overrightarrow{\mathrm{E}}+\overrightarrow{\mathrm{v}}\times \mathrm{}\overrightarrow{\mathrm{B}}]$

The **magnetic force** has following characteristics,

- It depends on q, v and B (charge of the particle, the velocity and the magnetic field). Force on a negative charge is opposite to that on a positive charge.
- The magnetic force
$$\overrightarrow{{\mathrm{F}}_{\mathrm{B}}}\mathrm{}=\mathrm{}\mathrm{q}\mathrm{}\mathrm{}\overrightarrow{\mathrm{v}}\times \mathrm{}\overrightarrow{\mathrm{B}}=\mathrm{}\mathrm{q}\mathrm{v}\mathrm{B}\mathrm{}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{\theta}\mathrm{}\widehat{\mathrm{n}}$$

includes a vector product of velocity and magnetic field. The vector product makes the force due to magnetic field zero if velocity and magnetic field are parallel or anti-parallel and maximum when they are at right angles to each other.

The force acts in a (sideways) direction perpendicular to both of them and Its direction is given by the screw rule or

**right hand rule**for vector (or cross) product**.**

- Only a moving charge feels the magnetic force.

**Magnetic force on a current carrying conductor**

Consider a rod of a uniform cross-sectional area *A* and length *L*. Let the number density of mobile charge carriers (electrons) in it be *n*. Then the total number of mobile charge carriers in it is *nLA*. For a steady current *I* in this conducting rod, each mobile carrier has an average drift velocity v_{d}.

In the presence of an external magnetic field B, the force on these carriers is:

$$\overrightarrow{{\mathrm{F}}_{\mathrm{B}}}=\mathrm{}\left(\mathrm{n}\mathrm{L}\mathrm{A}\right)\mathrm{q}\mathrm{}\overrightarrow{{\mathrm{v}}_{\mathrm{d}}}\mathrm{}\times \overrightarrow{\mathrm{}\mathrm{B}}$$

where q is the value of the charge on a carrier.

Now,

Current density* *$\overrightarrow{\mathrm{j}\mathrm{}}=\mathrm{}\mathrm{n}\mathrm{}\mathrm{q}\mathrm{}\overrightarrow{{\mathrm{v}}_{\mathrm{d}}}$

and

Current *I = |(n q v*_{d}*)| A*

Thus,

$$\overrightarrow{\mathrm{F}}\mathrm{}=\mathrm{}\left[\left(\mathrm{n}\mathrm{}\mathrm{q}\overrightarrow{\mathrm{}{\mathrm{v}}_{\mathrm{d}}}\mathrm{}\right)\mathrm{L}\mathrm{}\mathrm{A}\right]\times \mathrm{}\overrightarrow{\mathrm{}\mathrm{B}}\mathrm{}$$

$$=\mathrm{}\left[\mathrm{}\overrightarrow{\mathrm{j}\mathrm{}}\mathrm{A}\mathrm{}\mathrm{L}\right]\times \mathrm{}\overrightarrow{\mathrm{}\mathrm{B}}\mathrm{}=\mathrm{}\mathrm{I}\mathrm{}\overrightarrow{\mathrm{L}}\mathrm{}\times \mathrm{}\overrightarrow{\mathrm{}\mathrm{B}}$$

Or *F = B I L sin θ*

In this equation, B is the external magnetic field. It is not the field produced by the current-carrying rod.

For any arbitrary shape of the conductor,

$$\mathrm{F}\mathrm{}=\mathrm{}\mathrm{\Sigma}\mathrm{}\mathrm{I}\mathrm{}\mathrm{}\overrightarrow{\mathrm{d}\mathrm{L}}\mathrm{}\times \mathrm{}\overrightarrow{\mathrm{}\mathrm{B}}$$

This summation can be converted to an integral in most cases.

**Fleming’s left hand rule**

[Given direction of external magnetic field and current, find direction of force on it]

If we stretch the thumb, the forefinger and the central finger of left hand in such a way that all three are perpendicular to each other, then, if **Forefinger** represents the direction of magnetic **Field**, **Central finger** represents the direction of **Current** flowing through the conductor, then **thuMb** will represent the **direction of magnetic force (or Motion).**

**Motion of charge in a magnetic field – circular motion**

A force on a particle does work if the force has a component along (or opposed to) the direction of motion of the particle.

In the case of motion of a charge in a magnetic field, the magnetic force is **perpendicular** to the velocity of the particle. So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed).

The force due to an electric field, i.e., $\mathrm{q}\overrightarrow{\mathrm{E}}$, can have a component parallel (or antiparallel) to motion and thus can transfer energy in addition to momentum.

Consider motion of a charged particle in a **uniform** magnetic field.

Let us first take the case of v perpendicular to B. The perpendicular force, $\mathrm{q}\mathrm{v}\mathrm{B}$, acts as a centripetal force and produces a circular motion perpendicular to the magnetic field.

Now since *θ = 90*^{o}, If *r* is the radius of circular path, then,

$$\frac{\mathrm{m}{\mathrm{v}}^{2}}{\mathrm{r}}=\mathrm{}\mathrm{q}\mathrm{v}\mathrm{B}\mathrm{}$$

The radius of the circular path

$$\mathrm{}\mathrm{r}\mathrm{}=\frac{\mathrm{m}\mathrm{v}\mathrm{}}{\mathrm{B}\mathrm{q}\mathrm{}}$$

And

$$\mathrm{v}=\frac{\mathrm{q}\mathrm{B}\mathrm{r}}{\mathrm{m}}\mathrm{}$$

The larger the momentum (or velocity) the larger is the radius and bigger the circle described. If ω is the angular frequency, then

$$\mathrm{v}\mathrm{}=\mathrm{}\mathrm{\omega}\mathrm{r}$$

And

$$\mathrm{\omega}\mathrm{}=\mathrm{}2\mathrm{\pi}\mathrm{}\mathrm{\nu}\mathrm{}=\mathrm{}\frac{\mathrm{q}\mathrm{B}}{\mathrm{m}}$$

Time period,* *

$$\mathrm{T}=\frac{2\mathrm{\pi}}{\mathrm{\omega}}=\frac{2\mathrm{\pi}\mathrm{m}}{\mathrm{B}\mathrm{q}}$$

When a charged particle enters the magnetic field at any angle except 90°, it follows **helical path. **

In this case, the velocity has a component along $\overrightarrow{\mathrm{B}}$. This component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to $\overrightarrow{\mathrm{B}}$ is as before a circular one, thereby producing a **helical motion**.

The distance travelled by the charged particle in one time period due to component of velocity *v cos θ* (denoted by $\mathrm{v}}_{\left|\right|$), is called **pitch** of the path

$$\mathrm{p}\mathrm{}=\mathrm{}{\mathrm{v}}_{\left|\right|}\mathrm{}\mathrm{T}\mathrm{}\mathrm{}$$

$$=\mathrm{T}\mathrm{}\times \mathrm{}\mathrm{v}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{}\mathrm{\theta}\mathrm{}=\frac{\mathrm{}2\mathrm{\pi}\mathrm{m}\mathrm{}{\mathrm{v}}_{\left|\right|}}{\mathrm{q}\mathrm{}\mathrm{B}}$$

**Motion of charge in combined electric and magnetic fields**

**Velocity selector**

A charge *q *moving with velocity *v* in presence of both electric and magnetic fields experiences a force given by

$$\overrightarrow{\mathrm{F}}\mathrm{}=[\mathrm{}\mathrm{q}\overrightarrow{\mathrm{E}}\mathrm{}+\mathrm{}\mathrm{q}\overrightarrow{\mathrm{v}\mathrm{}}\times \mathrm{}\overrightarrow{\mathrm{B}}]\mathrm{}\equiv \mathrm{}{\overrightarrow{\mathrm{F}}}_{\mathrm{E}}\mathrm{}+\mathrm{}{\overrightarrow{\mathrm{F}}}_{\mathrm{B}}$$

- Consider the simple case in which electric and magnetic fields are perpendicular to each other and also perpendicular to the velocity of the particle.

We have,

$${\overrightarrow{\mathrm{F}}}_{\mathrm{E}}=\mathrm{q}\overrightarrow{\mathrm{E}}=\mathrm{q}\mathrm{E}\widehat{\mathrm{j}}\mathrm{}\mathrm{}$$

and

$$\mathrm{}{\overrightarrow{\mathrm{F}}}_{\mathrm{B}\mathrm{}}=\mathrm{}\mathrm{q}\overrightarrow{\mathrm{v}\mathrm{}}\times \mathrm{}\overrightarrow{\mathrm{B}}$$

$$=\mathrm{}\mathrm{q}\left(\mathrm{v}\mathrm{}\widehat{\mathrm{i}}\times \mathrm{B}\widehat{\mathrm{k}}\right)=-\mathrm{q}\mathrm{v}\mathrm{B}\widehat{\mathrm{j}}\mathrm{}$$

Therefore total force,

$$\overrightarrow{\mathrm{F}}=\mathrm{q}\left(\mathrm{E}-\mathrm{v}\mathrm{B}\right)\widehat{\mathrm{j}}$$

Thus, electric and magnetic forces are in opposite directions. If we adjust the value of E and B such that magnitudes of the two forces are equal, then, total force on the charge is zero and the charge will move in the fields **undeflected**.

Or

$$\mathrm{q}\mathrm{E}\mathrm{}=\mathrm{}\mathrm{q}\mathrm{v}\mathrm{B}\mathrm{}$$

$$\Rightarrow \mathrm{v}\mathrm{}=\frac{\mathrm{E}}{\mathrm{B}}$$

We can therefore select charged particles of a particular velocity out of a beam containing charges moving with different speeds (irrespective of their **charge** and **mass**).

The **crossed E and B fields**, therefore, serve as a **velocity selector**.

Only particles with speed $\frac{\mathrm{E}}{\mathrm{B}}$ pass undeflected through the region of crossed fields.

**CYCLOTRON**

Cyclotron is a device used to accelerate positively charged particles such as proton, deuteron etc.

**Structure of cyclotron**

Cyclotron consists of two large metallic D’s, connected to an oscillator which can change polarity of the D’s at regular intervals, a source of positively charged particles at the centre and a deflector plate as shown in the diagram.

**Principle of Cyclotron**

A positively charged particle can be accelerated through a moderate electric field by crossing it again and again by use of strong magnetic field.

The frequency *ν*_{a} of the applied voltage is adjusted so that the polarity of the Dees is reversed in the same time that it takes the ions to complete one half of the revolution.

The requirement *ν*_{a }*= ν*_{c} is called the **resonance condition**.

The phase of the supply is adjusted so that when the positive ions arrive at the edge of D_{1}, D_{2} is at a lower potential and the ions are accelerated across the gap.

Inside the Dees the particles travel in a region free of the electric field. The increase in their kinetic energy is *qV* each time they cross from one Dee to another (*V* refers to the voltage across the dees at that time).

The radius of their path goes on increasing each time their kinetic energy increases. The ions are repeatedly accelerated across the Dees until they have the required energy to have a radius approximately that of the Dees.

They are then deflected by a magnetic field and leave the system via an exit slit.

Radius of circular path

$$\mathrm{}\mathrm{r}\mathrm{}=\frac{\mathrm{m}\mathrm{v}}{\mathrm{B}\mathrm{q}}$$

Cyclotron frequency

$${\mathrm{\nu}}_{\mathrm{c}}\mathrm{}=\frac{\mathrm{B}\mathrm{q}}{2\mathrm{\pi}\mathrm{m}}$$

$$\Rightarrow \mathrm{}\mathrm{T}=\frac{2\mathrm{\pi}\mathrm{m}}{\mathrm{B}\mathrm{q}}$$

where *m* and *q* are mass and charge of the positive ion and *B* is strength of the magnetic field.

**Maximum kinetic energy gained by the particle in the cyclotron**

Now,

$$\mathrm{v}=\frac{\mathrm{q}\mathrm{B}{\mathrm{r}}_{\mathrm{o}}}{\mathrm{m}}$$

Hence,

$${\mathrm{E}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{1}{2}\mathrm{m}{\mathrm{v}}^{2}=\frac{{\mathrm{B}}^{2}{\mathrm{q}}^{2}{{\mathrm{r}}_{\mathrm{o}}}^{2}}{2\mathrm{m}}\mathrm{}$$

where, *r*_{o} = maximum radius of circular path.

- As the speed of ion becomes comparable with that of light, the mass of the ion increases according to the relation.
$$\mathrm{m}\mathrm{}=\frac{{\mathrm{m}}_{\mathrm{o}}}{\sqrt{1-\mathrm{}\frac{{\mathrm{v}}^{2}}{{\mathrm{c}}^{2}}}}$$

Where,

*m* = mass of the ion,

*m*_{o} = original mass of the ion

*v* = speed of ion and

*c* = speed of light.

**Limitations of Cyclotron**

- Cyclotron cannot accelerate uncharged particle like neutron.
- The positively charged particles having large mass i.e., ions cannot move at limitless speed in a cyclotron

**Force between two infinitely long parallel current carrying conductors**

Consider two long parallel conductors ‘*a*’ and ‘*b*’ separated by a distance *d* and carrying (parallel) currents *I*_{a} and *I*_{b}, respectively. The conductor ‘*a*’ produces, the same magnetic field *B*_{a} at all points along the conductor ‘*b*’ which is downwards.

$${\mathrm{B}}_{\mathrm{a}}\mathrm{}=\frac{\mathrm{\mu}\mathrm{o}}{2\mathrm{\pi}}\mathrm{}\frac{{\mathrm{I}}_{\mathrm{a}}}{\mathrm{d}}$$

The conductor ‘b’ carrying a current **I**_{b} will experience a sideways force due to the field B_{a}. The direction of this force is towards the conductor ‘a’.

$${\mathrm{F}}_{\mathrm{b}\mathrm{a}}\mathrm{}=\mathrm{}{\mathrm{I}}_{\mathrm{b}}\mathrm{}\mathrm{L}\mathrm{}{\mathrm{B}}_{\mathrm{a}}=\mathrm{}\frac{\mathrm{\mu}\mathrm{o}}{2\mathrm{\pi}}\mathrm{}\frac{{\mathrm{I}}_{\mathrm{a}}{\mathrm{I}}_{\mathrm{b}\mathrm{}}}{\mathrm{d}}\mathrm{L}$$

Similarly

$${\mathrm{F}}_{\mathrm{a}\mathrm{b}}\mathrm{}=\mathrm{}{\mathrm{I}}_{\mathrm{a}}\mathrm{}\mathrm{L}\mathrm{}{\mathrm{B}}_{\mathrm{b}}=\mathrm{}-\frac{\mathrm{\mu}\mathrm{o}}{2\mathrm{\pi}}\mathrm{}\frac{{\mathrm{I}}_{\mathrm{a}}{\mathrm{I}}_{\mathrm{b}\mathrm{}}}{\mathrm{d}}\mathrm{L}$$

We can see that *F*_{ab}* = - F*_{ba}

The force is attraction force if current in both conductors is in same direction and repulsion if current in both conductors is in opposite direction.

- If the currents in both parallel wires are equal and in same direction, then magnetic field at a point exactly half way between the wires is zero.

**Definition of Ampere**

The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2×10^{–7} newtons per metre of length.

**Defnition of Coulomb**

When a steady current of 1A is set up in a conductor, the quantity of charge that flows through its cross-section in 1s is one coulomb (1C).

**Torque acting on a current carrying coil placed inside a uniform magnetic field**

Consider the rectangular loop placed in such a way that the uniform magnetic field *B* is in the plane of the loop.

The field exerts no force on the two arms AD and BC of the loop. It is perpendicular to the arm AB of the loop and exerts a force *F*_{1} on it which is directed into the plane of the loop,

$${\mathrm{F}}_{1}\mathrm{}=\mathrm{}\mathrm{I}\mathrm{}\mathrm{b}\mathrm{}\mathrm{B}$$

Similarly it exerts a force *F*_{2} on the arm CD which is directed out of the plane of the paper,

$${\mathrm{F}}_{2}\mathrm{}=\mathrm{}\mathrm{I}\mathrm{}\mathrm{b}\mathrm{}\mathrm{B}\mathrm{}{\mathrm{F}}_{1}$$

Thus, the net force on the loop is zero. However there is a torque on the loop due to the pair of forces F_{1} and F_{2}.

$$\mathrm{\tau}\mathrm{}=\mathrm{}{\mathrm{F}}_{1}\times \frac{\mathrm{a}}{2}+\mathrm{}{\mathrm{F}}_{2}\times \frac{\mathrm{a}}{2}$$

$$=\mathrm{I}\mathrm{b}\mathrm{B}\mathrm{a}\mathrm{}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{}=\mathrm{}\mathrm{I}\mathrm{}\mathrm{a}\mathrm{b}\mathrm{}\mathrm{B}\mathrm{sin}\mathrm{}$$

*or τ = I A B sin θ*

Torque acting on a current carrying coil with *N* turns,

*τ = NBIA sin θ*

Where,

*N* = number of turns in the coil,

*B* = magnetic field intensity,

*I* = current in the coil,

*A* = area of cross-section of the coil and

*θ* = angle between magnetic field and normal to the plane of the coil.

**Current loop as magnetic dipole**

The current carrying loop behaves as a small magnetic dipole placed along the axis. One face of the loop behaves as north-pole while the other face of loop behaves as south-pole.

**The clock face rule**

While looking at the face of the coil, if the current is found to be flowing in the anti-clockwise direction, then the face of the coil will behave like the north pole and if the current is found to be flowing in the clockwise direction, the face of the coil will behave like south pole.

Each magnetic dipole has some magnetic moment (*m*). The vector $\overrightarrow{\mathrm{m}}$** **is given by

$$\overrightarrow{\mathrm{m}}\mathrm{}=\mathrm{}\mathrm{N}\mathrm{}\mathrm{I}\mathrm{}\overrightarrow{\mathrm{A}}$$

Hence

$$\overrightarrow{\mathrm{\tau}}\mathrm{}=\mathrm{}\overrightarrow{\mathrm{m}}\times \overrightarrow{\mathrm{B}}$$

This expression is similar to electrostatic case

$$\overrightarrow{\mathrm{\tau}}\mathrm{}=\mathrm{}\overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{E}}$$

SI unit of *m* are Am^{2}

**Stable Equilibrium**

When *m* and *B* are parallel.

**Unstable Eqilibrium**

When *m* and *B* are anti-parallel.

**Difference between electric dipole and magnetic dipole**

Electrical Dipole is combination of two elementary charges, Charges can exist individually. However no magnetic monopoles exist.

**The magnetic dipole moment of a revolving electron**

Ampere suggested that all magnetism is due to circulating currents. This seems to be partly true and no magnetic monopoles have been seen so far. However, elementary particles such as an electron or a proton also carry an **intrinsic** magnetic moment, not accounted by circulating currents.

Let us consider an electron that is revolving around in a circle of radius *r* with a velocity *v*.

The electron of charge (– *e*) (*e* = + 1.6 × 10^{–19} C) performs uniform circular motion around a stationary heavy nucleus of charge +*Ze*.

Current I = $\frac{\mathrm{e}}{\mathrm{T}}$

*T* is the time period of revolution. Let *r* be the orbital radius of the electron, and *v* the orbital speed, then,

$$\mathbf{}\mathrm{T}\mathrm{}=\frac{2\mathrm{\pi}\mathrm{r}}{\mathrm{v}}$$

$$\Rightarrow \mathrm{I}=\frac{\mathrm{e}\mathrm{v}}{2\mathrm{\pi}\mathrm{r}}$$

The magnetic moment, denoted by* μ*_{l}, associated with this circulating current is,

$$\mathrm{}{\mathrm{\mu}}_{\mathrm{l}}\mathrm{}=\mathrm{}\mathrm{I}\mathrm{\pi}{\mathrm{r}}^{2}\mathrm{}=\frac{\mathrm{e}\mathrm{v}\mathrm{r}}{2}$$

Multiplying and dividing the right-hand side of the above expression by the electron mass *m*_{e}, we have,

$${\mathrm{\mu}}_{\mathrm{l}}\mathrm{}=\mathrm{}\frac{\mathrm{e}\left({\mathrm{m}}_{\mathrm{e}}\mathrm{v}\mathrm{r}\right)}{2\mathrm{}{\mathrm{m}}_{\mathrm{e}}}=\frac{\mathrm{e}\mathrm{l}}{2{\mathrm{m}}_{\mathrm{e}}},\mathrm{}$$

Where *l* = angular momentum of the electron

- The ratio, $\frac{\mathrm{e}\mathrm{l}}{2{\mathrm{m}}_{\mathrm{e}}}$ is called the
**gyromagnetic ratio**^{10 }C/kg for an electron, which has been verified by experiments. - Bohr predicted that the angular momentum of an electron can only take values
*l*= $\frac{\mathrm{n}\mathrm{h}}{2\mathrm{\pi}}$, hence,$${\mathrm{}\mathrm{\mu}}_{\mathrm{l}}\mathrm{}=\mathrm{}\frac{\mathrm{e}}{2{\mathrm{m}}_{\mathrm{e}}}\times \frac{\mathrm{n}\mathrm{h}}{2\mathrm{\pi}}$$

Putting

*n*= 1, we get the minimum magnetic moment (*μ*_{l})_{min}= 9.27 × 10^{–24}Am^{2}. This is called**Bohr Magneton.** - This dipole moment is labeled as the orbital magnetic moment. Hence the subscript ‘
*l*’ in*μ*_{l}. Besides the orbital moment, the electron has an intrinsic magnetic moment, which has the same numerical value as orbital magnetic moment. It is called the spin magnetic moment. - But the electron is not spinning. The electron is an elementary particle and it does not have an axis to spin around like a top or our earth.
- The microscopic roots of magnetism in iron and other materials can be traced back to this intrinsic spin magnetic moment.

**Moving coil galvanometer**

It is a device used for the detection and measurement of the currents. It works on the principle of conversion of electrical energy into mechanical energy. When a current flows through the coil, a torque acts on it.

*τ = N I A B*

In equilibrium, deflecting torque = restoring torque, or

$$\mathrm{k}\mathrm{}=\mathrm{}\mathrm{N}\mathrm{I}\mathrm{A}\mathrm{B}\mathrm{}\mathrm{}\mathrm{}\Rightarrow \mathrm{}\mathrm{}\mathrm{}=\frac{\mathrm{N}\mathrm{A}\mathrm{B}}{\mathrm{k}}\mathrm{I}$$

Hence θ ∝ I

where,

k = restoring torque per unit twist,

N = number of turns in the coil,

B = magnetic field intensity,

A = area of cross-section of the coil and

θ = angle of twist.

**Q:** Why Galvanometer cannot be used to measure current?

- Galvanometer is a very sensitive device, it gives a full-scale deflection for a current of the order of
*μA*. - For measuring currents, the galvanometer has to be connected in series, and as it has a large resistance, this will change the value of the current in the circuit.

**Ammeter - Shunt resistance**

To overcome these difficulties, a small resistance r_{s}, called **shunt resistance**, is attached in parallel with the galvanometer coil; so that most of the current passes through the shunt.

Since, *R*_{G}* >> r*_{s}*,* the resistance of this arrangement is,

$$\frac{{\mathrm{R}}_{\mathrm{G}}{\mathrm{}\mathrm{r}}_{\mathrm{s}}\mathrm{}}{{\mathrm{R}}_{\mathrm{G}}\mathrm{}+\mathrm{}{\mathrm{}\mathrm{r}}_{\mathrm{s}}}\simeq \mathrm{}{\mathrm{}\mathrm{r}}_{\mathrm{s}}$$

**Current sensitivity of galvanometer**

The deflection produced per unit current in galvanometer is called its current sensitivity. Current sensitivity

$${\mathrm{I}}_{\mathrm{s}}\mathrm{}=\frac{\mathrm{\theta}}{\mathrm{I}}=\frac{\mathrm{N}\mathrm{B}\mathrm{A}}{\mathrm{k}}$$

**Voltmeter**

The galvanometer can be used as a voltmeter. For this it must be connected in parallel with that section of the circuit. It must draw a very small current, otherwise the voltage measurement will disturb the original set up by an amount which is very large. Hence a large resistance R is connected **in series** with the galvanometer.

The voltmeter resistance is,

*R*_{G}* + R *$\simeq$* R (very large)*

**Voltage sensitivity of galvanometer**

The deflection produced per unit voltage applied across the ends of galvanometer is called its voltage sensitivity.

$$\mathrm{V}\mathrm{s}\mathrm{}=\frac{\mathrm{\theta}}{\mathrm{V}}=\frac{\mathrm{N}\mathrm{B}\mathrm{A}}{\mathrm{k}\mathrm{R}}$$

**For a sensitive galvanometer**

- N should be large
- B should be large
- A should be large
- k should be small

Q: Increasing the current sensitivity may not necessarily increase the voltage sensitivity. Explain.

If N → 2N, i.e., we double the number of turns, then current sensitivity will double.

However, the resistance of the galvanometer is also likely to double, since it is proportional to the length of the wire (Number of turns).

N → 2N will lead to R → 2R, thus the voltage sensitivity, remains unchanged.