Magnetic Effect of Current Class 12 Physics | Notes

Magnetic effect of current

When current flows through a conductor, a magnetic field is developed around it. This effect of current is known as magnetic effect of current.

Orested discovery

In 1820 AD, Orested established a relation between electricity and magnetism. The experimental setup for Orested experiment is as shown in figure. In figure, a conducting wire is connected with a battery through an ammeter, a key and a variable resistance. A compass needle is placed parallels to the length of wire. When circuit is opened, the compass needle does not show any deflection and points north and south. When the current is passed through the conductor, the compass needle shows deflection and when the direction of current is reversed the compass needle shows deflection in opposite direction. Thus the magnetic field is developed around a current carrying conductor.

Magnetic Effect of Current Class 12 Physics | Notes

Representation of magnetic field.

(i) Horizontal magnetic field

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(ii) Oblique magnetic field

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(iii) Magnetic field perpendicularly inward to the plane of paper.

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(iv) Magnetic field perpendicularly outward to the plane of paper.

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Right hand thumb rule: 

This rule is applicable for current carrying straight conductor. According to right hand thumb rule, if the thumb of right hand represents the direction of current of straight conductor then the tip of curled fingers encircling the conductor will give the direction of magnetic field.

Magnetic Effect of Current Class 12 Physics | Notes

 Right hand first rule:

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This rule is used to find the direction of magnetic field due to a current carrying a circular coil. 

According to this rule, a current carrying circular coil is gripped in the right hand in such a way that the curved fingers in the direction of current.  On doing so, the thump gives direction to the magnetic field.

Fleming’s Left hand rule:

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This rule is used to find direction of force on a charged particle moving in a magnetic field.

According to this rule, the fore finger,  middle finger and thumb of left hand are stressed mutually perpendicular to each other in such a way that the fore finger in the direction of magnetic field and middle finger in the direction of motion of charged then the thump gives direction of force on the charged particle.

Biot and Savart Law (1831 A.D)

Biot and Savart Law Is used to find magnitude of magnetic field due to a current carrying conductor of any shape and size.
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Let us consider a conductor carrying current (I). Let us take a small element of length (dl) at point (O).  We have to find the magnetic field at point (p) due to the small element (dl).

I.e. OP = r

In figure, angle between (dl) and (r) is $\theta $o. According to Biot and Savart law, the magnetic field at point (P) due to the small element (dl) is:

i) directly proportional to magnitude of current (i.e. dB $\propto $ I) ……………….(i)

ii) directly proportional to length of a small element (i.e. dB $\propto $ dl) ……………….(ii)

iii) directly proportional to sine of the angle between dl and r (i.e. dB $\propto $ sin$\theta $) ……………….(iii)

iv) inversely proportional to the square of distance (i.e. dB $\propto $ $\frac{I}{{{r}^{2}}}$) ……………….(iv)

Now,

On combining equation (i), (ii), (iii) & (iv)

i.e.  $dB\text{ }\propto \text{ }\frac{I\text{ }dl\text{ }\sin \theta }{{{r}^{2}}}$

$dB\text{  =  }\frac{K\text{ }I\text{ }dl\text{ }\sin \theta }{{{r}^{2}}}$ ………..(v) {Where K is proportionality constant & its value is found to be $\frac{{{\mu }_{0}}}{4\pi }$where ${{\mu }_{0}}$ is magnetic permeability of free of space and its value in SI system is $4\pi \times {{10}^{-7}}H{{m}^{-1}}$}

In vector form, the above result can be written as:

$\left[ \overrightarrow{dB}\text{ = }\frac{{{\mu }_{0}}}{4\pi }\text{ I  }\left( \frac{\overrightarrow{dl}\text{ }\times \text{ }\overrightarrow{r}\text{ }}{{{\left| \overrightarrow{r} \right|}^{3}}} \right)\text{  } \right]$

In CGS System, equation (v) can be written as

$\left[ dB\ =\ \frac{I\ dl\ \sin \theta }{{{r}^{2}}} \right]$

Application of Biot and Savart law

1. Magnetic field due to a current carrying circular coil.

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Let us take a circular coil of radius (r) and Centre at (O) and carrying current (I). We have to calculate magnetic field at point (O) i.e. at centre of circular coil due to the current carrying circular coil. Let us take a small element of length (dl) at point (P). Let us join (O) and (P) i.e. OP = r. The angle between dl and ‘r’ is${{90}^{{}^\circ }}$.

According to Biot and service law, the magnetic field at point O due to the small element (dl) is

i.e. 

dB = $\frac{{{\mu }_{0}}}{4\pi }$$$$\frac{I\text{ }dl\text{ }\sin \theta }{{{r}^{2}}}$

dB = $\frac{{{\mu }_{0}}}{4\pi }$ $\frac{I\text{ }dl\text{ }\sin {{90}^{{}^\circ }}}{{{r}^{2}}}\text{  }\left( \because \theta ={{90}^{{}^\circ }} \right)$

dB =$\frac{{{\mu }_{0}}}{4\pi }$ $\frac{I\text{ }dl}{{{r}^{2}}}$……….(i)

Now, 

Total magnetic field due to the circular coil can be obtained by integrating equation (i) from limit 0 to 2$\pi $r.

i.e. B=$\int\limits_{0}^{2\pi r}{dB}\text{ }=\text{ }\frac{{{\mu }_{0}}I}{4\pi {{r}^{2}}}\int\limits_{0}^{2\pi r}{dl}$

B = $\frac{{{\mu }_{0}}I}{4\pi {{r}^{2}}}\text{ }2\pi r$

B = $\frac{{{\mu }_{0}}I}{2r}$ $$$$

For N – Turns  $\left[ B\text{ }=\text{ }\frac{{{\mu }_{0}}\text{ }NI}{2r}\text{ } \right]$

The direction of magnetic field is along the axis of circular coil.

1. A wire of length 62.8m Carrying current 10A is bent into a circular coil of radius 10cm.  10 cm magnet at centre. 

Soln,

l = 62.8m

r = 10cm = 0.1m

I = 10A

B =?

Now,

N = $\frac{l}{2\pi r}=\frac{62.8}{2\times 3.14\times 0.1}=100$

Now,

B = $\frac{{{\mu }_{0}}NI}{2r}$

= $\frac{4\pi \times {{10}^{-7}}\times 100\times 10}{2\times 0.1}$

=$6.28\times {{10}^{-3}}$T.

2. An electron is revolving around its orbit of radius $5\times {{10}^{-11}}$m with a frequency of$2\times {{10}^{10}}$Hz. Calculate magnitude of magnetic field at its centre.

Soln,

r = $5\times {{10}^{-11}}$m

f = $2\times {{10}^{10}}$Hz

Now, I = $\frac{e}{t}$   =  ef    =   $1.6\times {{10}^{-19}}\times 2\times {{10}^{10}}$

= $3.2\times {{10}^{-9}}$A

Now, 

B = $\frac{{{\mu }_{0}}I}{2r}$

= $\frac{4\pi \times {{10}^{-7}}\times 3.2\times {{10}^{-9}}}{2\times 5\times {{10}^{-11}}}$

= $4.02\times {{10}^{-5}}\,$T.

2. Magnetic field along the axis of narrow circular coil.

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Let us take on narrow circular coil of radius (a) centered at (O) and carrying current (I) in anti-clockwise direction. We have to calculate magnetic field at point (P) along x-axis at a distance (X) from centre (O) i.e. OP = X. Let us take small element of length (dl) at point (C).  Let us join C and P i.e. CP = r. The angle between dl & r is 90o.

According to Biot & Savart law, magnetic field at point (P) due to the small element (dl),

dB = $\frac{{{\mu }_{0}}}{4\pi }$ $\frac{I\text{ }dl\text{ }\sin \theta }{{{r}^{2}}}$

dB = $\frac{{{\mu }_{0}}}{4\pi }$ $\frac{I\text{ }dl\text{ }\sin 90{}^\circ }{{{r}^{2}}}\left( \because \text{ }\theta \text{ = 90}{}^\circ  \right)$

dB = $\frac{{{\mu }_{0}}}{4\pi }$ $\frac{I\text{ }dl}{{{r}^{2}}}$……….(i)

(Direction along PQ)

On resolving dB into its constituent components, we get $dB\cos \alpha $ and $dB\sin \alpha $ as shown in figure.

Let us take another small element X’Y’ (dl) at diametrically opposite to XY. 

Since all the components $dB\cos \alpha $ being equal in magnitude, acting in opposite direction cancel to each other. But all the components $dB\sin \alpha $are being equal in magnitude and acting in same direction are added up.

Now, total magnetic field can be obtained by integrating either $dBdin\alpha $ from limit 0 to 2$\pi $a.

 $B\text{ = }\int\limits_{0}^{2\pi a}{dB\sin \alpha }$

$B\text{ = }\int\limits_{0}^{2\pi a}{\frac{{{\mu }_{{}^\circ }}}{4\pi }\text{ }\frac{I\text{ }dl}{{{r}^{2}}}\text{ }\frac{a}{r}\text{ }\left( \because \sin \alpha \text{ = }\frac{a}{r}\text{ } \right)}$

$B\text{ = }\frac{{{\mu }_{0}}\text{ }I\text{ a}}{4\pi {{r}^{3}}}\ \text{ }\int\limits_{0}^{2\pi a}{dl}$

$B\text{ = }\frac{{{\mu }_{0}}\text{ }I\text{ a}}{4\pi {{r}^{3}}}\ \text{ }2\pi a$

$B\text{ = }\frac{{{\mu }_{0}}\text{ }I\ {{a}^{2}}}{2{{r}^{3}}}$

$B\text{ = }\frac{{{\mu }_{0}}\text{ }I\text{ }{{\text{a}}^{2}}}{2{{\left( {{a}^{2\text{ }}}+\text{ }{{\text{x}}^{2}} \right)}^{\frac{3}{2}}}}$      $\left( \because \text{ r = }\sqrt{{{a}^{2}}\text{ + }{{\text{x}}^{2}}} \right)$

For N- turns

$B\text{ = }\frac{{{\mu }_{0}}\text{ N }I\text{ }{{\text{a}}^{2}}}{2{{\left( {{a}^{2\text{ }}}+\text{ }{{\text{x}}^{2}} \right)}^{\frac{3}{2}}}}$

Special Cases:

1. If x $\rangle \rangle \rangle $ a, in this case the value of a can be neglected as compared to x.

We have,

$B\text{ = }\frac{{{\mu }_{0}}\text{ N }I\text{ }{{\text{a}}^{2}}}{2{{\left( \text{ }{{\text{x}}^{2}} \right)}^{\frac{3}{2}}}}$

$\therefore \text{ }B\text{ = }\frac{{{\mu }_{0}}\text{ N }I\text{ }{{\text{a}}^{2}}}{2{{x}^{3}}}$

2. If a >>x or x=0

In this case, the value of x can be neglected as compared to a. 

We have, 

$B\text{ = }\frac{{{\mu }_{0}}\text{ N }I\text{ }{{\text{a}}^{2}}}{2{{\left( \text{ }{{\text{a}}^{2}} \right)}^{\frac{3}{2}}}}$

$B\text{ = }\frac{{{\mu }_{0}}\text{ N }I\text{ }{{\text{a}}^{2}}}{2\text{ }{{\text{a}}^{3}}}$

$\therefore \text{ }B\text{ = }\frac{{{\mu }_{0}}\text{ N }I\text{ }}{2\text{ a}}$

3. Magnetic Field due to a current carrying straight conductor:

Magnetic Effect of Current Class 12 Physics | Notes

Let us take a straight conductor, XY carrying current (I) in upward direction. We have to calculate magnetic field at point (P) at a distance (a) from the straight conductor i.e. OP = a. Let us take small element of length (dl) at point (C) i.e. OC = l 

Let us join C and P i.e. CP = r

In figure, $\measuredangle \text{ CPO = }\phi $and $\measuredangle \text{ PCO = }\theta $ 

According to Biot and Savart law, the magnetic field at point P due to small element the (dl) at point (C) is

$dB\text{ = }\frac{{{\mu }_{0}}}{4\pi }\text{ }\frac{I\text{ d}l\text{ sin}\theta }{{{r}^{2}}}$……………………(i)

From

$\text{sin}\theta \text{ =}\frac{a}{r}\text{ = cos}\phi $

Also r = a sec$\phi $

Again,

$\tan \phi \text{ = }\frac{l}{a}$

l =$\text{ a tan}\phi $

Differentiating both side with respect to $\phi $

$dl\text{ = a se}{{\text{c}}^{2\text{ }}}\phi \text{ d}\phi $

Now, equation (i) becomes

$dB\text{ = }\frac{{{\mu }_{0}}}{4\pi }\text{ }\frac{I\text{ a Se}{{\text{c}}^{2}}\phi \text{ d }\phi \text{ cos }\phi }{{{a}^{2}}\text{ }{{\sec }^{2}}\text{ }\phi }$

$\therefore \text{ dB  = }\frac{{{\mu }_{0}}I}{4\pi \text{ a}}\text{ cos }\phi \text{ d }\phi $……………(ii)

Now, total magnetic field can be obtained by integrating equation (ii) from limit $-\text{ }{{\phi }_{1}}$ to ${{\phi }_{2}}$.

B = $\text{ }\int_{-{{\phi }_{1}}}^{{{\phi }_{2}}}{dB}$

Or, B = $\frac{{{\mu }_{o}}I}{4\pi a}$$\int_{-{{\phi }_{1}}}^{{{\phi }_{2}}}{\cos \phi d\phi }$

Or, B = x$\frac{{{\mu }_{o}}I}{4\pi a}$$\left| \sin \phi  \right|_{-{{\phi }_{1}}}^{{{\phi }_{2}}}$

Or, B = $\frac{{{\mu }_{o}}I}{4\pi a}$[sin${{\phi }_{2}}$ – sin($-\text{ }{{\phi }_{1}}$)]

Or, B = $\frac{{{\mu }_{o}}I}{4\pi a}$(sin${{\phi }_{2}}$ + sin${{\phi }_{1}}$)

Or, B = $\frac{{{\mu }_{o}}I}{4\pi a}$(sin${{\phi }_{1}}$ + sin${{\phi }_{2}}$)

The above result gives magnetic field due to a current carrying straight conductor of finite length. If the conductor is infinitely long then

 ${{\phi }_{1}}$= ${{\phi }_{2}}$= $\frac{\pi }{2}$

Or, B = $\frac{{{\mu }_{o}}I}{4\pi a}$(sin$\frac{\pi }{2}$ + sin$\frac{\pi }{2}$)

Or, B = $\frac{{{\mu }_{o}}I}{4\pi a}$(1+1)

Or, B = $\frac{{{\mu }_{o}}I}{2\pi a}$

The above result gives magnetic field due to a current carrying straight conductor of infinite length.

4. Magnetic field due to a long current carrying solenoid.

A long cylindrical coil having large number of turns is known as solenoid.
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Let us consider a long solenoid of total number of turns N, number of turns per unit length ‘n’ and carrying current I in anti-clockwise direction. Let radius of solenoid be ‘a’. In figure each circle represents one turn of solenoid. We have to calculate magnetic field at point ‘P’ inside the solenoid. Let us consider an elementary length ‘dl’ and ‘C’ any point on it. Let us join C and P. i.e. CP = r. 

In figure CQ = a

OP = lo  

OQ = l

PQ = lo l

According to Biot- Savart law, magnetic field at point ‘P’ due to the small element ‘dl’ is,

dB = $\frac{{{\mu }_{o}}I{{a}^{2}}}{2{{r}^{3}}}\times N’$

Where $N’$is the number of turns in small length ‘dl’.

Let ‘n’ be the number of turns per unit length.

Then n = $\frac{N’}{dl}$

$\therefore $ $N’$= ndl

dB = $\frac{{{\mu }_{o}}I{{a}^{2}}}{2{{r}^{3}}}\times ndl$…………(i)

From $\Delta CPQ$

Or, sin$\theta $= $\frac{a}{r}$

Or, r =$\frac{a}{\sin \theta }$= a cosec$\theta $

Again,

tan$\theta $= $\frac{CQ}{PQ}$= $\frac{a}{{{l}_{o}}-l}$

Or, lo l = a cot$\theta $

Differentiating above equation w.r.t. $\theta $

Or, – dl = $-a\cos e{{c}^{2}}\theta \text{ }d\theta $

Or,  dl = $a\cos e{{c}^{2}}\theta \text{ }d\theta $

Now equation (i) becomes,

Or,  dl = $\frac{{{\mu }_{o}}I{{a}^{2}}}{2{{a}^{3}}\cos e{{c}^{3}}\theta }\times n$$a\cos e{{c}^{2}}\theta \text{ }d\theta $

Or,  dl = $\frac{{{\mu }_{o}}nI}{2}$$\sin \theta \text{ }d\theta $…….(ii)

Now the total magnetic field can be obtained by integrating equation (ii) from $-{{\theta }_{1}}$to${{\theta }_{2}}$ 

B = $\text{ }\int_{-{{\theta }_{1}}}^{{{\theta }_{2}}}{dB}$= $\frac{{{\mu }_{o}}nI}{2}$$\int_{-{{\theta }_{1}}}^{{{\theta }_{2}}}{\sin \theta \text{ }d\theta }$

B = $\frac{{{\mu }_{o}}nI}{2}$$\left| -\cos \theta  \right|_{{{\theta }_{1}}}^{{{\theta }_{2}}}$

B = $\frac{{{\mu }_{o}}nI}{2}$[-cos$\theta {}_{2}$-(-cos(-$\theta {}_{1}$))]

B = $\frac{{{\mu }_{o}}nI}{2}$[-cos$\theta {}_{2}$+ cos$\theta {}_{1}$]

B = $\frac{{{\mu }_{o}}nI}{2}$(cos$\theta {}_{1}$- cos$\theta {}_{2}$)

The above result gives magnetic field due to a long solenoid of finite length. if the solenoid is infinitely long then

$\theta {}_{1}$= 0o and $\theta {}_{2}$= $\pi $

B = $\frac{{{\mu }_{o}}nI}{2}$(cos0- cos$\pi $)

B = $\frac{{{\mu }_{o}}nI}{2}$(1+1)

B = ${{\mu }_{o}}nI$

N.B.

The above result gives magnetic field inside a long solenoid is almost uniform.

The magnetic field at the one end of solenoid is 

B = $\frac{{{\mu }_{o}}nI}{2}$

Ampere’s Circuital law: (is not universal)

It states that “The line integral of magnetic field in a close loop is equal μo times the total current enclosed by that loop”. 

i.e. $\oint{\overrightarrow{B}}.\overrightarrow{dl}$ =  ${{\mu }_{o}}I$

Ampere’s circuital law is used to find magnetic field due to current carrying conductor. 

i.e., objective of Ampere’s circuital law is same as Biot and Savart law. 

Ampere’s circuital law is much easier and faster than Biot and Savart law but it can be applied only for symmetrical conductors.

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Proof: 

Let us take a long straight conductor XY carrying current I. Let any point P at a distance ‘a’ from the conductor i.e. OP = a. According to Biot and Savart law, the magnetic field at point P due to the current carrying conductor XY is 

B  =  $\frac{{{\mu }_{o}}I}{2\pi a}$

Let us draw a close loop from point P concentric with conductor XY. Let us take a small element of length dl at point P. The direction of the magnetic field is given by tangent PQ. i.e. direction of $\overrightarrow{B}$ and $\overrightarrow{dl}$ is same. So the angle between them is 0°. 

Now, line integral of magnetic field in close loop is 

i.e. $\oint{\overrightarrow{B}}.\overrightarrow{dl}$= $\oint{{}}$B dl cos$\theta $

= $\oint{{}}$ B dl cos 0°

= B   $\int\limits_{0}^{2\pi a}{{}}$d l

= $\frac{{{\mu }_{o}}I}{2\pi a}$ .2$\pi $a

= ${{\mu }_{o}}$I

$\therefore $ $\oint{\overrightarrow{B}}.\overrightarrow{dl}$=  ${{\mu }_{o}}$ I

Application of Ampere’s circuital law

1. Magnetic field due to a current carrying straight conductorC:\Users\user\AppData\Local\Temp\Rar$DIa0.821\CamScanner 04-02-2021 21.26_2.jpg

Let us take a long straight conductor XY carrying current I. We have to calculate magnetic field at point P at a distance ‘a’ from the straight conductor i.e. OP = a. Let us draw a close loop from point P concentric with a straight conductor. Let us take a small element of length dl at point P. The direction of magnetic field at point P is given by tangent PQ. i.e. direction of $\overrightarrow{B}$ and $\overrightarrow{dl}$ are the same. So, the angle between them is 0°. 

From Ampere’s circuital law,

i.e. $\oint{\overrightarrow{B}}.\overrightarrow{dl}$ =  ${{\mu }_{o}}$ I

or, $\oint{{}}$ Bd l cos $\theta $ = ${{\mu }_{o}}$I

or, $\oint{{}}$ Bd l cos 0° = ${{\mu }_{o}}$I

or, B $\int\limits_{0}^{2\pi a}{{}}$  d l = ${{\mu }_{o}}$I

or, B . 2$\pi $a = ${{\mu }_{o}}$I

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

2. Magnetic field due to a long solenoid.

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Let us consider a long solenoid of total number of turns ‘N’, no. of turns per unit length ‘n’ and carrying current ‘I’ by each turn of the solenoid. In figure, each circle represents each turn of the solenoid. We have to calculate the magnetic field inside the long solenoid. The magnetic field inside the long solenoid is almost uniform and directed along the axis but the magnetic field outside the solenoid is so minimal that can be assumed to be zero. 

Let us consider a rectangular closed loop PQRSP of length PQ = dl,

No. of turns in elementary length dl =$N’$

No. of turns per unit length, n= $\frac{N’}{dl}$

$\therefore $ $N’$ = ndl

From Ampere’s circuital law,

i.e. $\oint{\overrightarrow{B}}.\overrightarrow{dl}$=  ${{\mu }_{o}}$ ×Itotal

     $\oint{\overrightarrow{B}}.\overrightarrow{dl}$=  ${{\mu }_{o}}$ I×$N’$

     $\oint{\overrightarrow{B}}.\overrightarrow{dl}$=  ${{\mu }_{o}}$ I×ndl…………. (i) 

Now the line integral of magnetic field,

 $\oint{\overrightarrow{B}}.\overrightarrow{dl}$=  $\int\limits_{P}^{Q}{{}}$ $\overrightarrow{B}$ . $\overrightarrow{PQ}$  +  $\int\limits_{Q}^{R}{{}}$ $\overrightarrow{B}$ . $\overrightarrow{QR}$ + $\int\limits_{R}^{S}{{}}$ $\overrightarrow{B}$ . $\overrightarrow{RS}$  + $\int\limits_{S}^{P}{{}}$ $\overrightarrow{B}$ . $\overrightarrow{SP}$       …………. (ii) 

Now,

$\oint{\overrightarrow{B}}.\overrightarrow{dl}$=     $\int\limits_{P}^{Q}{{}}$  B dl Cos0o + $\int\limits_{Q}^{R}{{}}$ B dl Cos90o   + $\int\limits_{R}^{S}{{}}$B dl Cos180o   +$\int\limits_{S}^{P}{B}$  dl Cos270o     

$\int\limits_{Q}^{R}{{}}$ B dl Cos90o   = 0 $\because $ Cos90o =0

 $\int\limits_{R}^{S}{{}}$B dl Cos180o = 0 ($\because $B = 0 outside solenoid)

$\int\limits_{S}^{P}{B}$  dl Cos270o = 0 ($\because $ Cos270o =0)

So equation (ii) becomes,   

$\oint{\overrightarrow{B}}.\overrightarrow{dl}$=$\int\limits_{P}^{Q}{{}}$ B dl +0+0+0

$\oint{\overrightarrow{B}}.\overrightarrow{dl}$= B dl

From equation (i)

B dl  = ${{\mu }_{o}}$ I×ndl

$\therefore $B  =  ${{\mu }_{o}}$nI

.3. Magnetic field due to a toroid. 

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Let us consider a toroid of total no. of turns ‘N’, no. of turns per unit length ‘n’ and carrying current ‘I’ in anticlockwise direction. Let ‘r’ be the mean radius of the toroid and it’s center is at O. We have to calculate the magnetic field inside the toroid. Let us draw a closed loop concentric with toroid represented by dotted lines as shown in the figure. Let us consider a small element of length d l . The direction of $\overrightarrow{B}$ and $\overrightarrow{dl}$ are the same. So, the angle between them is 0°.

No. of turns per unit length = n = $\frac{N}{2\pi r}$

$\therefore $N = n 2$\pi $r

From Ampere’s circuital law,

i.e. $\oint{\overrightarrow{B}}.\overrightarrow{dl}$ =  ${{\mu }_{o}}$ ITotal

      $\oint{{}}$Bd l cos $\theta $ = ${{\mu }_{o}}$I N

      $\oint{{}}$Bd l cos 0° =${{\mu }_{o}}$I  n 2$\pi $r

      B $\int\limits_{0}^{2\pi r}{dl}$ = ${{\mu }_{o}}$I n 2$\pi $r

       B  2$\pi $r = ${{\mu }_{o}}$n 2$\pi $rI

$\therefore $ B   =  ${{\mu }_{o}}$nI

Force on a charged particle moving in a uniform magnetic field. (Lorentz force)

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Let a charged particle of magnitude ‘q’ enters in uniform magnetic field B with velocity V making an angle $\theta $ with the direction of magnetic field. When a charged particle enters in a magnetic field, it experiences a force which is known as Lorentz force. 

Experimentally, it is found that the Lorentz force experienced by charged particle is 

i. Directly proportional to magnitude of magnetic field. 

i.e.  F $\propto $ B  ………………….. (i) 

ii. Directly proportional to magnitude of charge.

i.e.  F $\propto $ q  ……………. (ii) 

iii. Directly proportional to velocity of charged particle, 

i.e.  F $\propto $ V  ……………. (iii)

iv. Directly proportional to sine of the angle between velocity and magnetic field. 

F $\propto $ sin$\theta $ ……………….. (iv) 

On combining equation (i), (ii), (iii), (iv); we get;

F $\propto $ BqV  sin$\theta $

F = K BqV sin$\theta $

Where K is proportionality constant whose value is found to be 1.

F = Bqv sin $\theta $

In vector form;

$\overrightarrow{F}$ = q ($\overrightarrow{V}$ × $\overrightarrow{B}$) force is 1 to velocity and magnetic field.

Special cases:

1. If $\theta $ = 0°  or  180°

We have,

F  =  BqV sin$\theta $

F  =  BqV sin 0     [$\because $   $\theta $  =  0°]

F  =  0

The above result shows that magnetic field does not exert any force on the charged particle when it is moving in

the direction or in the opposite direction of the magnetic field. 

2. If $\theta $ = 90°

We have,

F  =  BqV sin$\theta $

F  =  BqV sin 90°   [$\because $   $\theta $  =  90°]

F  =  BqV

The above result shows that charged particle experience maximum force when it is moving in a direction

perpendicular to the magnetic field. 

3. If q = 0

We have,

F  =  BqV sin$\theta $

F  =  0   [$\because $   q  =  0]

The above result shows that the magnetic field does not exert any force on the neutral particle. 

Force on a current carrying conductor placed in a uniform magnetic field. 

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Let us consider a conductor of length l, area of cross section ‘A’ and carrying current ‘I’ is placed in uniform magnetic field ‘B’ in such a way that the plane of conductor makes an angle $\theta $ with the direction of magnetic field. 

Since a conductor contains a large number of free electrons even at normal temperature. Let ‘n’ be the no. of electrons per unit volume and ‘e’ be the charge on each electron. The current flowing in the conductor means the free electrons are moving in the opposite direction with certain velocity known as drift velocity (V).

Let total no. of electrons = N 

Volume of conductor = Al

 No. of electrons per unit volume, n = $\frac{N}{Al}$

$\therefore $Total no. of electrons N = nAl

We know that when an electron moves in a magnetic field, it experiences a force which is known as Lorentz force. The Lorentz force experienced by each electron is 

FL  =  BeV sin $\theta $

Total force experienced by conductor 

F = NFL

F = N BeV sin$\theta $  [$\because $   FL= BeV sin $\theta $]

F = nAl BeV sin$\theta $  [$\because $   N = nAl]

F = BVenA l sin$\theta $

F = BIl sin$\theta $  [$\because $I  =  VenA]

Special Cases

1. If $\theta $ = 0

F  =  BI l sin $\theta $

F  =  BI l sin 0°

F  =  0

The above result shows that a current carrying conductor placed in magnetic field does not experience any force

when it’s plane is in the direction or opposite direction of the magnetic field. 

2. If $\theta $ = 90°

We have,

F  =  BI l sin$\theta $

F  =  BI l sin 90°

F  =  BI l

The above result shows that a current carrying conductor placed in a magnetic field experiences maximum force when it is plane is perpendicular to the direction of magnetic field. 

Force between parallel current carrying conductors:

When current flows through a conductor, a magnetic field will be developed around it. When a current carrying conductor is placed inside a magnetic field it experiences force. When two parallel current carrying conductors are placed close to each other, it is equivalent that one of the current carrying conductors is inside the magnetic field of another conductor. So, there exists a mutual force between them. The force may be attractive or repulsive depending upon direction of current. 

1. When current flows in same direction (Like current) 

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Let two conductors X and Y of length l1 and l2 carrying current I1 and I2 in the same direction are placed

parallel to each other separated by a distance ‘a’.

According to Biot and Savart law, the magnetic field at any point P of conductor Y due to current carrying conductor X is 

B1  = $\frac{{{\mu }_{0}}{{I}_{1}}}{2\pi a}$

Now, the current carrying conductor Y lies inside the magnetic field B1 of conductor X. The force experienced by conductor Y is 

${{F}_{2}}’$ =  B1I2 l2

or, ${{F}_{2}}’$ =  $\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}{{l}_{2}}}{2\pi a}$

Force per unit length is:

F2 =  $\frac{{{F}_{2}}’}{{{l}_{2}}}$  

or, F2 =  $\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{2\pi a}$ ………………… (i)

The direction of F2 is towards point Q 

According to Biot and Savart law, magnetic field at any point Q of conductor X due to current carrying conductor Y is 

B2 =  $\frac{{{\mu }_{0}}{{I}_{2}}}{2\pi a}$

Now,

the current carrying conductor X lies inside the magnetic field B2 of conductor Y.

Now, force experienced by current carrying conductor X is

${{F}_{1}}’$ =  B2 I1 l1

${{F}_{1}}’$= $\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}{{l}_{1}}}{2\pi a}$

Force per unit length is 

F1 =  $\frac{{{F}_{1}}’}{{{l}_{1}}}$  

F1 = $\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{2\pi a}$  ………………… (ii)

The direction of F1 is towards point P.

From equation (i) and (ii) 

$\therefore $F1 =  F2  =$\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{2\pi a}$

This force is attractive in nature. 

2. When current flows in opposite direction (Unlike current)

C:\Users\user\Desktop\CamScanner 03-28-2021 21.41_20.jpg

Similarly,

F1  =   F2  = $\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{2\pi a}$

This force is repulsive in nature. 

Define: 1 Ampere current in terms of force between parallel current carrying conductors. 

The force between parallel current carrying conductors carrying current I1 and I2 separated by a distance a is

F  = $\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{2\pi a}$

If I1  =  I2  =  1A and a = 1 m then, 

F  =  $\frac{4\pi \times {{10}^{-7}}\times 1\times 1}{2\pi \times 2}$

$\therefore $ F  =  2 × 10–7 Nm–1

Current is said to be of 1A if two parallel conductors carrying equal amounts of current separated by unit distance experience 2 × 10–7 Nm–1 force per unit length. 

Torque on a rectangular coil suspended in a uniform magnetic field.

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Let a rectangular coil PQRS of length PQ = RS = l and breadth QR = SP = b, carrying current. I in anticlockwise direction is suspended in uniform magnetic field B. Suppose at an instant of time the plane of coil makes an angle $\theta $ with the direction of magnetic field i.e. the breadth wise position of the coil makes an angle $\theta $ with magnetic field but length wise position is always perpendicular to the direction of magnetic field. 

When current flows through the rectangular coil, it’s each arm experience force. The force experienced by each arm is given below: 

1. Force on PQ

$\overrightarrow{{{F}_{1}}}$ = I  ($\overrightarrow{PQ}$×$\overrightarrow{B}$)

or, $\overrightarrow{{{F}_{1}}}$ = I  ($\overrightarrow{l}$ ×$\overrightarrow{B}$)

or, F1 = BIl sin$\theta $

or, F1 = BIl sin 90°

or, F1 = BIl  ……………….. (i) 

2. Force on arm QR

$\overrightarrow{{{F}_{2}}}$ = I ($\overrightarrow{QR}$ × $\overrightarrow{B}$)

or, $\overrightarrow{{{F}_{2}}}$ = I ($\overrightarrow{b}$ × $\overrightarrow{B}$)

or, F2  =  BIb sin$\theta $ ……………. (ii)

3. Force on arm RS

$\overrightarrow{{{F}_{3}}}$ = I ($\overrightarrow{RS}$ × $\overrightarrow{B}$)

or, $\overrightarrow{{{F}_{3}}}$  =  I($\overrightarrow{l}$ × $\overrightarrow{B}$)

or, F3  =  BIl sin$\theta $

or, F3  =  BIl sin 90°

or, F3  =  BI l ………………… (iii)

4. Force on arm SP

$\overrightarrow{{{F}_{4}}}$ = I ($\overrightarrow{SP}$ × $\overrightarrow{B}$)

or, $\overrightarrow{{{F}_{4}}}$= I ($\overrightarrow{b}$ × $\overrightarrow{B}$)

or, F4  =  BIb sin(180 – $\theta $)

or, F4  =  BIb sin$\theta $  …………….. (iv) 

The force F2 and F4 being equal in magnitude acting in opposite directions with the same line of action cancel to each other but the forces F1 and F3 being equal in magnitude acting in opposite directions with different line of action constitute a couple. 

The torque due to couple of force F1 and F3 is

Torque ($\tau $)  =  Force on either arm × Perpendicular distance
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or, $\tau $  =  F1 × PT

or, $\tau $  =  F1PS cos$\theta $  [⸪   From $\Delta $PTS cos$\theta $ = $\frac{PT}{PS}$]

or, $\tau $  =  BI l b cos$\theta $   [⸪   PS = b]

or, $\tau $  =  BIA cos$\theta $

For N-turns

$\therefore $$\tau =BINA\cos \theta $

Special cases

1. If θ = 0°

We have,  $\tau =BINA\cos \theta $ 

or, $\tau $  =  BINA cos 0°

or, $\tau $  =  BINA (Max. torque)

The above result shows that if plane of coil is parallel to the direction of magnetic field, it experience maximum torque. 

2. If $\theta $ = 90°,

  $\tau =BINA\cos \theta $

or, $\tau $  =  BINA cos 90°

or, $\tau $  =  0  (No torque)

The above result shows that if plane of coil is perpendicular to the direction of magnetic field, it exp. no. torque. 

Moving coil galvanometer

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Moving coil galvanometer

It is an electrical instrument which is used to measure small amount of current and potential difference. 

Principle

A moving coil galvanometer works upon the principle that if current is passed through a rectangular coil suspended in uniform magnetic field, it experience torque. The torque experienced by the rectangular coil is. 

$\tau =BINA\cos \theta $

Construction

A moving coil galvanometer consist of a rectangular coil ABCD made up of insulated copper wire. The insulated copper wire wound on non-magnetic metallic frame with soft iron as a central core. The rectangular coil is suspended from torsion head with the help of phosphor bronze stripe in between concave pole piece of permanent magnet so that the magnetic field becomes radial. The lower end of coil is connected with hair spring made up of quartz. Concave mirror maximize image. So, used. Plane give same size. Convex decrease. Concave mirror is placed on the phosphor bronze strip as shown in figure. The terminals T1 and T2 act as path of current. 

A lamp-scale arrangement is placed to note the perfect deflection. 

Working

When current is passed through the rectangular coil ABCD suspended in radial magnetic field with the help of phosphor bronze stripe experience deflecting torque due to which the coil gets deflected. As the coil deflected the phosphor bronze strip twisted from it’s original position. When the phosphor bronze strip twisted, a restoring torque develop in it which tries to bring the strip towards it’s original position. The twisting of phosphor bronze strip takes place until an equilibrium is established between restoring torque and deflecting torque. Suppose at equilibrium condition the strip twisted by an angle $\theta $.

Now, the restoring torque on the strip is

Restoring torque (${{\tau }_{R}}$)  =  c$\theta $  ……………… (i)

Where c is torsion constant. 

The deflecting torque experienced by the rectangular coil of total number of turns N, area A, carrying current I suspended in radial magnetic field B is

Deflecting torque (${{\tau }_{d}}$)  =  BINA cos$\theta $ …………….. (ii)

Since the magnetic field is radial. i.e. plane of coil is always parallel to the direction of magnetic field. So, $\theta $ = 0°. 

Now, Equation (ii) becomes; 

Deflecting torque (${{\tau }_{d}}$)  =  BINA cos 0°

or,     ${{\tau }_{d}}$  =  BINA  ………………………. (iii)

At equilibrium condition, 

${{\tau }_{d}}$  =  ${{\tau }_{R}}$

or, BINA  =  C$\theta $

or, I  =  $\frac{C\theta }{BNA}$

$\therefore $ I  =  G$\theta $

Where G  =  $\frac{C}{BNA}$ is constant and known as galvanometer constant. 

1. Current sensitivity

It is defined as the deflection per unit current. 

We have, 

C$\theta $ =  BINA

or, $\frac{\theta }{I}$ =  $\frac{BINA}{c}$

$\therefore $ Current sensitivity, SI =  $\frac{BNA}{c}$

2. Voltage sensitivity

It is defined as the deflection per unit potential difference.

We have,

C$\theta $  =  BINA

or, $\frac{\theta }{I}$  =  $\frac{BNA}{c}$

or, $\frac{\theta }{IR}$  =  $\frac{BNA}{cR}$

or, $\frac{\theta }{V}$  =  $\frac{BNA}{cR}$

$\therefore $ Voltage sensitivity, SV  =  $\frac{BNA}{cR}$

Hall’s Effect

Hall’s Effect states that if a magnetic field is applied on a current carrying conductor in a direction perpendicular to that of current, an electric field will be developed which is perpendicular to direction of magnetic field and electric current.C:\Users\user\Downloads\CamScanner 04-01-2021 21.56.jpg

Let us take a rectangular conductor of thickness ‘t’ and breadth ‘d’. Since the conductor contains a large number of free electrons even at normal temperature. Let ‘n’ be the no. of electrons per unit volume and ‘e’ be the charge on each electron. Let the conductor be connected with a battery in such a way that current (Ix) flows in positive X direction. It means that electrons are moving with certain velocity (Vx) known as drift velocity along negative X direction. Now, a magnetic field is applied along the Z direction. Due to this magnetic field (Bz), the moving electron experiences a force known as Lorentz force. The Lorentz force experienced by each electron is

Fl = Bz eV………………….(i)

Due to this Lorentz force, the free electrons shifted in upward direction leaving an equal amount of positive charge in downward direction. Due to this force electrons are accumulated in upward direction and protons are accumulated in downward direction. Due to this difference in charge and electric field (Ex) will be set up on the conductor along Y direction. This field is known as Hall field and potential due to the Hall field is known as Hall potential.

FE =  eE………………….(ii)

The accumulation process of electrons takes place until an equilibrium is established between Lorentz force and electrostatic force.

At equilibrium,

FE = FL………………………(iii)

From equation (i), (ii) and (iii)

eEH  = Bz evx  

or, EH  = Bz v………………….(iv)

From expression of drift velocity,

Ix = VxenA

or, Vx = $$$\frac{{{I}_{x}}}{e\text{ }n\text{ A}}$……………………..(v)

${{E}_{H\text{ }}}=\text{ }{{\text{B}}_{Z\text{ }}}\frac{{{I}_{x}}}{e\text{ n A}}$

or, $\frac{{{V}_{H}}}{d}\text{ = }\frac{{{B}_{z}}{{I}_{x}}}{en\left( d\times t \right)}$

$\therefore $ ${{V}_{H\text{ }}}=\text{ }\frac{{{B}_{Z}}{{I}_{x}}}{n\text{ e t}}$

This is the required expression for Hall voltage.

Expression for Hall coefficient

Hall’s Effect states that if a magnetic field is applied on a current carrying conductor in a direction perpendicular to that of current, an electric field will be developed which is perpendicular to direction of magnetic field and electric current.C:\Users\user\Downloads\CamScanner 04-01-2021 21.56.jpg

Let us take a rectangular conductor of thickness ‘t’ and breadth ‘d’. Since the conductor contains a large number of free electrons even at normal temperature. Let ‘n’ be the no. of electrons per unit volume and ‘e’ be the charge on each electron. Let the conductor be connected with a battery in such a way that current (Ix) flows in positive X direction. It means that electrons are moving with certain velocity (Vx) known as drift velocity along negative X direction. Now, a magnetic field is applied along the Z direction. Due to this magnetic field (Bz), the moving electron experiences a force known as Lorentz force. The Lorentz force experienced by each electron is

Fl = Bz eV………………….(i)

Due to this Lorentz force, the free electrons shifted in upward direction leaving an equal amount of positive charge in downward direction. Due to this force electrons are accumulated in upward direction and protons are accumulated in downward direction. Due to this difference in charge and electric field (Ex) will be set up on the conductor along Y direction. This field is known as Hall field and potential due to the Hall field is known as Hall potential.

FE =  eE………………….(ii)

The accumulation process of electrons takes place until an equilibrium is established between Lorentz force and electrostatic force.

At equilibrium,

FE = FL………………………(iii)

From equation (i), (ii) and (iii)

eEH  = Bz evx  

or, EH  = Bz v………………….(iv)

From expression of current density

${{J}_{X\text{ }}}=\text{ – n e }{{\text{v}}_{x}}$

or, ${{v}_{x}}\text{ = – }\frac{{{J}_{x}}}{n\text{ e}}$…………………….(v)

From equation (iv) and (v)

${{E}_{H}}\text{  }{{\text{B}}_{Z}}\text{ }\left( -\text{ }\frac{{{J}_{x}}}{n\text{ e}} \right)$

or, $-\frac{1}{n\text{ e}}\text{ = }\frac{{{E}_{H}}}{{{B}_{z}}{{J}_{x}}}$

$\therefore $$$${{R}_{H\text{  }}}=\text{ }\frac{{{E}_{H}}}{{{B}_{z}}{{J}_{x}}}$

Where, ${{R}_{H}}\text{ = – }\frac{1}{n\text{ e}}$  is constant and known as the Hall coefficient.

Also read: Heating Effect of Current

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