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

If the self inductance of 500 turns coil is 125 mH, then the self inductance of the similar coil of 800 mH

  1. 48.8 mH

  2. 200 mH

  3. 290 mH

  4. 320 mH
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$L=\mu _o \mu _r N^2Al$
For similar coil, $A, l$ will be same 
So,   $ \, \dfrac{L _1}{L _2} = \dfrac{N _1^2}{N _2^2}$

$ L _{800}= \dfrac{N _{800} ^2}{N _{500}^2}\times L _{500}= \, \dfrac{125}{(500)^2} \, \times \, (800)^2 \, = \, 320 \, mH$

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

The mutual inductance $M _{12}$ of a coil 1 with respect to coil 2

  1. increases when they are brought nearer

  2. depends on the current passing through the coils.

  3. increases when one of them is rotated about an axis.

  4. both (a) and (b) are correct
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mutual Induction: Whenever the current passing through a coil or circuit changes, the magnetic flux linked with a neighbouring coil or circuit will also change. Hence an emf will be induced in the neighbouring coil or circuit. This phenomenon is called ‘mutual induction’.


If the two coils $1$ and $2$ are present with mutual inductance $M _1$ and $M _2$. Then the mutual inductance of the coil 1 due to 2 increases when they are bought near since, mutual inductance is  proportional to the flux passed through the coil.

The mutual induction of $M _{12}$ is same as $M _{21}$

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

Match the following:

Quantity Formula
1) Magnetic flux linked with a coil a) $\displaystyle -N\frac { d\phi  }{ dt } $
2) Induced emf b) $\displaystyle { \mu  } _{ r }{ \mu  } _{ 0 }{ n } _{ 1 }{ n } _{ 2 }{ \pi r } _{ 1 }^{ 2 }l$
3) Force on a charged particle moving in a electric and magnetic field c) $\displaystyle BA\cos { \theta  } $
4) Mutual inductance of a solenoid d) $\displaystyle q\left( \overline { E } +\overline { v } \times \overline { B }  \right) $

  1. 1-c, 2-d, 3-b, 4-a

  2. 1-c, 2-a, 3-d, 4-b

  3. 1-b, 2-a, 3-c, 4-d

  4. 1-a, 2-b, 3-d, 4-c
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

1) Magnetic flux through any area is the scalar product of its area vector with the magnetic field vector. Thus for a coil, it is $\vec{B}.\vec{A}=BAcos\theta$

2) Emf induced in a coil due to changing flux through it is given by Faraday's Law,
$Emf = -N\dfrac{d\phi}{dt}$
3) Force on a charged particle due to electric field = $q\vec{E}$
Force on a moving charged particle due to magnetic field = $q(\vec{v}\times \vec{B})$
Thus, force on a moving charged particle in an electric and magnetic field = $q(\vec{E}+\vec{v}\times \vec{B})$
4) Mutual inductance of a solenoid is found out to be : $\mu _r\mu _0n _1n _2\pi r _1^2l$

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

In the method using the transformers, assume that the ratio of the number of turns in the primary to that in secondary in the step-up transformer is $1:10$. If the power to the consumer has to be supplied at $200\ V$, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is:

  1. $200:1$

  2. $150:1$

  3. $100:1$

  4. $50:1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

 An inductor of inductance $100\ mH$ is connected in series with a resistance, a variable capacitance and an AC source of frequency $2.0\ kHz$; The value of the capacitance so that maximum current may be drawn into the circuit. 

  1. 50 nF

  2. 60 nF

  3. 63 nF

  4. 79 nF
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l}{X _L} = Lw = {10^{ - 1}} \times 2\pi  \times 2 \times {10^3}\{X _L} = 4\pi  \times {10^2}\Z = \sqrt {{{\left( {{X _L} - {X _C}} \right)}^2} + {R^2}} \i = \dfrac{V}{Z} = \dfrac{V}{{\sqrt {{{\left( {{X _L} - {X _C}} \right)}^2} + {R^2}} }}\for,{i _{\max }}\{X _L} = {X _C}\\therefore {X _C} = Lw = \dfrac{1}{{Cw}}\C = \dfrac{1}{{{w^2}L}} = \dfrac{1}{{{{10}^{ - 1}} \times 4{\pi ^2} \times 4 \times {{10}^6}}}\ = \dfrac{{{{10}^{ - 5}}}}{{16{\pi ^2}}} = 63nF\end{array}$

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

In mutual induction 
A: when current in one coil increases, induced current in neighbouring coil flows in the opposite direction
B: When current in one coil decreases, induced current in neighbouring coil flows in the opposite direction

  1. A is true, B is false

  2. A and B are false

  3. A and B are true

  4. A is false, B is true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Lenz's Law states that the induced current will flow in a direction that opposes the change in magnetic flux. If current in the primary coil increases, the induced current in the secondary coil creates a magnetic field opposing the increase (opposite direction). If current decreases, the induced current creates a field to support the flux (same direction). Thus, A is true and B is false.

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

Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of $10^{-3}$ Wb to link with A and a flux per turn of $0.8 \times 10^{-3}$ Wb through B. The ratio of self-inductance of A and the mutual inductance of A and B is :

  1. 5/4

  2. 1/1.6

  3. 1.6

  4. 1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of 10−3 Wb to link with A and a flux per turn of 0.8×10−3 Wb through B. The ratio of self-inductance of A and the mutual inductance of A and B is $\dfrac{L _1}{L _2}=\dfrac{200*10^{-3}}{400*0.8*10^{-3}}=1/1.6$