# Tag: physics

### Questions Related to physics

The mutual inductance of an induction coil is 5 H. In the primary coil, the current reduces from 5 A to zero in $10^{-3} s$. What is the induced e.m.f. in the secondary coil?

1. 2500 V

2. 25000 V

3. 2510 V

4. zero

Correct Option: B
Explanation:

$EMF=L\dfrac { di }{ dt }$

$=5\times \dfrac { 5 }{ { 10 }^{ -3 } }$

$=25000V$

Two concentric rings are kept in the same plane. Number of turns in each rings is $25$. Their radii are $50 cm$ and $200 cm$ and they carry electric currents of $0.1 A$ and $0.2 A$ respectively, in mutually opposite directions. The magnitude of the magnetic field produced at their centre is ____________ $T$.

1. $2{ \mu } _{ 0 }$

2. $4{ \mu } _{ 0 }$

3. $\dfrac { 10 }{ 4 } { \mu } _{ 0 }$

4. $\dfrac { 5 }{ 4 } { \mu } _{ 0 }$

Correct Option: D
Explanation:

Given, ${ N } _{ 1 }={ N } _{ 2 }=25$ turns
${ R } _{ 1 }=50 cm=0.5 m$
${ R } _{ 2 }=200 cm=2 m$
${ i } _{ 1 }=0.1A, { i } _{ 2 }=0.2A$
The magnitude of the magnetic field
$\Delta B={ B } _{ 1 }-{ B } _{ 2 }$
$=\dfrac { { \mu } _{ 0 }{ N } _{ 1 }{ i } _{ 1 } }{ 2{ R } _{ 1 } } -\dfrac { { \mu } _{ 0 }{ N } _{ 2 }{ i } _{ 2 } }{ 2{ R } _{ 2 } }$
$=\dfrac { { \mu } _{ 0 }\times 25 }{ 2 } \left( \dfrac { { i } _{ 1 } }{ { R } _{ 1 } } -\dfrac { { i } _{ 2 } }{ { R } _{ 2 } } \right)$
$=\dfrac { { \mu } _{ 0 }\times 25 }{ 2 } \left( \dfrac { 0.1 }{ 0.5 } -\dfrac { 0.2 }{ 2 } \right)$
$=\dfrac { 25 }{ 2 } { \mu } _{ 0 }\left( \dfrac { 1 }{ 5 } -\dfrac { 1 }{ 10 } \right)$
$=\dfrac { 25 }{ 2 } { \mu } _{ 0 }\left( \dfrac { 2-1 }{ 10 } \right)$
$=\dfrac { 25 }{ 2 } { \mu } _{ 0 }\times \dfrac { 1 }{ 10 } =\dfrac { 25 }{ 20 } { \mu } _{ 0 }=\dfrac { 5 }{ 4 } { \mu } _{ 0 }$

A solenoid of length 30 cm with 10 turns per centimetre and area of cross-Section 40 $cm^2$completely surrounds another co-axial solenoid of same length, area of Cross-section 20 $cm^2$ with 40 turns per centimetre. The mutual inductance of the

1. 10 H

2. 8 H

3. 3mH

4. 30 mH

Correct Option: C
Explanation:

Given:

$n _1 = 10cm^{-1} = 1000 m^{-1}$
$n _2 = 40cm^{-1} = 4000 m^{-1}$
$l= 30cm = 30 \, \times \, 10m^{-2}$
$A _2 = 20cm^2 = 20 \, \times \, 10^{-4}m^2$

Mutual inductance of the system,
$M \, = \, \mu _0n _1n _2A _2l$                                  (Where $A _2$ is the area of inner solenoid.)

$\therefore M = 4\pi \times 10^{-7} \times 1000 \times 4000 \times 20 \times 10^{-4} \times 30 \times 10^{-2}$

$M= 301.44 \, \times 10^{-5} H = 3mH$

A 2 m long solenoid with diameter 2 cm an 2000 turns has a secondary coil of 1000 turns wound closely near its midpoint. The mutual inductance between the two coils is

1. $2.4\times 10^{-4}$ H

2. $3.9\times 10^{-4}$ H

3. $1.28\times 10^{-3}$ H

4. $3.14\times 10^{-3}$ H

Correct Option: B
Explanation:

Here, $l = 2 m$, diameter = 2 cm
$\therefore \, radius, \, r = \dfrac{2}{2} = 1cm = 1 \, \times \,10^{-2}m$

$N _1$ = 2000, $N _2$ = 1000

Area = $\pi r^2$  =  $\pi \, \times \, \left ( 1\times 10^{-2} \right )^2 = 3.14 \times 10^{-4} m^2$

Mutual inductance, M = $\dfrac{\mu _0N _1N _2A}{l}$

$\,= \, \dfrac{4\pi \,\times \,10^{-7}\, \times \, 2000 \, \times \, 1000 \, \times \, 3.14 \,\times \,10^{-4}}{2}$

$M=3.9 \, \times \, 10^{-4}H$

A pair of adjacent coils has a mutual inductance of 2.5 H. If the current in one coil changes from 0 to 40 A in 0.8 s, then the change in flux linked with the other coil is then

1. 100 Wb

2. 120 Wb

3. 200 Wb

4. 250 Wb

Correct Option: A
Explanation:

Mutual inductance of a pair of coils,, $M=2.5\ H$

Initial current, $i _1=0\ A$

Final current, $i _2=40\ A$

Change in current, $di=i _2-i _1=40\ A$

Time taken for the change, $t= 0.8\ sec$

Induced e.m.f, $e= \dfrac{d\phi}{dt}= M\dfrac{di}{dt}$

where $d\phi$ is the change in the flux linked with the coil.

$\implies d\phi = Mdi =2.5\times 40 =100\ Wb$

Hence, the change in the flux linkage is $100\ Wb$.

So,  option $(A) is correct. A short solenoid of radius a, number of turns per Unit length$n _1$. and length L is kept coaxially inside a very long solenoid of radius b, the number of turns per Unit length$n _2$. What is the mutual inductance of the system? 1.$\mu _0\pi b^2 n _1 n _2 L $2.$\mu _0\pi a^2 n _1 n _2 L ^2$3.$\mu _0\pi a^2 n _1 n _2 L $4.$\mu _0\pi b^2 n _1 n _2 L ^2$Correct Option: C Explanation: Let$L$be the length of each solenoid$S _1$and$S _2$having radius a and b respectively.$n _1$and$n _2$be the number of turns per unit length of$S _1$and$S _2$. And$I$be the current through solenoid$S _2$Magnetic field in$S _2= B _2= {\mu _0n _2I}$Magnetic flux linked with each turn of$S _1 = B _2 \times $area of each turn$= B _1\pi a^2$Total magnetic flux linked with$S _1= B _2\pi a^2n _1L\therefore \phi _1 = \left({\mu _0n _2I}\right)\pi a\ ^2n _2L = {\mu _0n _1n _2 \pi a^2 I}{L}$But magnetic flux linked with$S _1$is due to$I\therefore \phi _1 \propto I$or$\phi _1 = M\ I$Where$M$is the mutual inductance of$S _2$and$S _1\therefore M \, I = {\mu _0n _1n _2\pi a^2I}{L}\therefore M = \mu _0n _1n _2\pi a^2L$Two short bar magnets of magnetic moment 'M' each are arranged at the opposite corners of a square of side 'o', such that their centres coincide with the square. If the like poles are in the same direction, the magnetic induction at any of the other of the square is 1.$\frac { { \mu } _{ 0 } }{ 4\pi } \frac { M }{ { d }^{ 3 } } $2.$\frac { { \mu } _{ 0 } }{ 4\pi } \frac { 2M }{ { d }^{ 3 } } $3.$\frac { { \mu } _{ 0 } }{ 2\pi } \frac { 3M }{ { d }^{ 3 } } $4.$\frac { { \mu } _{ 0 } }{ 2\pi } \frac { 2M }{ { d }^{ 3 } } $Correct Option: A A conducting ring of radius r and resistance R rolls on a horizontal surface with constant velocity v. 1. The induced emf.between O and Q is 2 Bvr. 2. An induced current$ I= \dfrac { 2Bvr }{ R } $flows in the clockwise direction. 3. An induced current$ I= \dfrac { 2Bvr }{ R } $flows in the anticlockwise direction. 4. No current flows Correct Option: D Two coils of self-inductances$2\ mH$and$8\ mH$are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is: 1.$10\ mH$2.$6\ mH$3.$4\ mH$4.$16\ mH$Correct Option: C Explanation: Mutual inductance M=$\sqrt{ L _1\times L _2}=\sqrt{ 2\times 8}=4$mH Mutual inductance of a system of two thin coaxial conducting loops of radius each, their centers separated by distance$d (d >>r)$is 1.$\mu _0\pi r^4d^3$2.$\dfrac{\mu _0\pi r^4}{2d^3}$3.$\dfrac{\mu _0\pi r^4}{d^3}$4.$\dfrac{\mu _0\pi r^4 d^3}{4}\$

Correct Option: B