Tag: electromagnetic induction

Quizzes (6) Questions (197)

Questions Related to electromagnetic induction

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