Questions Related to physics

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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

Reveal answer Fill a bubble to check yourself
D Correct answer
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) $
           $=\dfrac { 25 }{ 2 } { \mu  } _{ 0 }\left( \dfrac { 2-1 }{ 10 }  \right)$
           $=\dfrac { 25 }{ 2 } { \mu  } _{ 0 }\times \dfrac { 1 }{ 10 } =\dfrac { 25 }{ 20 } { \mu  } _{ 0 }=\dfrac { 5 }{ 4 } { \mu  } _{ 0 }$

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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

Reveal answer Fill a bubble to check yourself
C Correct answer
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.)


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

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

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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

Reveal answer Fill a bubble to check yourself
B Correct answer
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 = $\dfrac{\mu _0N _1N _2A}{l}$

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

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

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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

Reveal answer Fill a bubble to check yourself
A Correct answer
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. 

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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. $\mu _0\pi b^2 n _1 n _2 L ^2$

Reveal answer Fill a bubble to check yourself
C Correct answer
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 $S _1$
$\therefore M \, I = {\mu _0n _1n _2\pi a^2I}{L}$

$\therefore M = \mu _0n _1n _2\pi a^2L$

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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 } } $

Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When a conducting ring rolls on a surface, the motional emf induced in the top half cancels the emf induced in the bottom half relative to the center, or the symmetry of the rolling motion results in zero net current in the ring.

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

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. $\dfrac{\mu _0\pi r^4 d^3}{4}$

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The mutual inductance between two small coaxial loops separated by distance d is M = (mu_0 * pi * r1^2 * r2^2) / (2 * d^3). Given r1 = r2 = r, M = (mu_0 * pi * r^4) / (2 * d^3).

Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

The coefficient of mutual inductance, when magnetic flux changes by $\displaystyle 2\times { 10 }^{ -2 }Wb$ and current changes by 0.01 A is :

  1. 8 henry

  2. 4 henry

  3. 3 henry

  4. 2 henry

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that

$\displaystyle \phi =Mi$

$\displaystyle d\phi =Mdi$

$\displaystyle M=\frac { d\phi  }{ di } =\frac { 2\times { 10 }^{ -2 } }{ 1\times { 10 }^{ -2 } } =2$ henry