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

Two concentric coils each of radius equal to $2\pi\ cm$ are placed at right angles to each other. $3$ Ampere and $4$ ampere are the currents flowing in each coil respectively. The magnetic induction in $Weber/m^{2}$ at the centre of the coils will be ($\mu _{0}=4\pi \times 10^{-7}\ Wb/A-m$)

  1. $12\times 10^{-5}$

  2. $10^{-5}$

  3. $5\times 10^{-5}$

  4. $7\times 10^{-5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Magnetic Induction at centre of coil is
$B=\dfrac {\mu _0}{2r} \sqrt {I _1^2 +I _2^2}\quad I _1=3\ A$
$I _2=4\ A$
$=\dfrac {4\pi\times 10^{-7}}{2\times \dfrac {2\pi}{100}}\times \sqrt {3^2 +4^2}$
$=\dfrac {4\pi \times 10^{-5}\times 5}{2\times 2\pi}$
$=5\times 10^{-5} \dfrac {wb}{m^2}$
Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

A 60 volt - 10 watt bulb is operated at 100 volt - 60 Hz a.c. The inductance required is?

  1. 2.56 H

  2. 0.32 H

  3. 0.64 H

  4. 1.28 H
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

First, find the resistance of the bulb: R = V^2 / P = 60^2 / 10 = 360 ohms. The operating current is I = P / V = 10 / 60 = 1/6 A. When connected to 100V, the impedance Z = V_source / I = 100 / (1/6) = 600 ohms. Since Z^2 = R^2 + Xl^2, 600^2 = 360^2 + Xl^2. Xl^2 = 360000 - 129600 = 230400. Xl = 480 ohms. Since Xl = 2 * pi * f * L, 480 = 2 * 3.14 * 60 * L. L = 480 / 377 = 1.273 H, which rounds to 1.28 H.

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

Two coaxial coils are very close to each other and their mutual inductance is $5mH$. If a current $50sin{500t}$ is passed in one of the coils then the peak value of induced emf in the secondary coil will be

  1. $5000V$

  2. $500V$

  3. $150V$

  4. $125V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

According to principle of mutual inductance, flux induced in coil  is equa to the current flowing in coil 1

$\phi _{2}= Mi _{1}$
By Faraday's laws,
$\dfrac{d\phi _{2}}{dt}=M\dfrac{di _{1}}{dt}$ = EMF
$\therefore EMF = 5\times 10^{-3}\dfrac{d}{dt}50 sin 500t$
$\therefore EMF = 125 cos500t$
So, maximum vaue of EMF would be 125 V

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

The coefficient of self induction of two inductor coils are $20mH$ and $40mH$ respectively. If the coils are connected in series so as to support each other and the resultant inductance is $80mH$ then the value of mutual inductance between the coils will be

  1. $5mH$

  2. $10mH$

  3. $20mH$

  4. $40mH$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$L _{total} = L _1 + L _2 + 2M$


$80 mH = 20 mH + 40 mH + 2M$

$\therefore M = 10 mH$

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

A rectangular loop of sides 'a' and 'b' is placed in the XY plane. A very long wire is also placed in xy plane such that side of length 'a' of the loop is parallel to the wire. The distance between the wire and the nearest edge of the loop is 'd'. The mutual inductance of this system is proportional to?

  1. a

  2. b

  3. $1/d$

  4. Current in wire
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The mutual inductance M between a long wire and a rectangular loop is calculated by integrating the magnetic flux through the loop. The flux is proportional to the dimensions of the loop, specifically the side length 'a' parallel to the wire, because the magnetic field B varies with distance from the wire. Thus, M is proportional to 'a'.

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

A circular loop of radius $r$ is placed at the centre of current carrying conducting square loop of side $a$. If both loops are coplanar and $a >> r$, then the mutual inductance between the loops will be:

  1. $\dfrac{\mu _0r^2}{2\sqrt{2}(a)}$

  2. $\dfrac{\mu _0r^2}{4a}$

  3. $\dfrac{2\sqrt{2}\mu _0r^2}{\pi a}$

  4. $\dfrac{\mu _0r^2}{4\sqrt{2}a}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Both loops are coplanar. 

Magnetic field at the center of outer square current carrying loop is
${ B } _{ 1 }=\dfrac { 2\sqrt { 2 } { \mu  } _{ 0 }I }{ \pi a } $
where $a$= length of side of square loop and
$r$= radius of the circular loop.
Given $a>>r$,
The magnetic field through entire inner coil is ${ B } _{ 1 }$
Magnetic flux through inner coil, ${ \phi  } _{ 21 }={ B } _{ 1 }{ A } _{ 2 }$
        =$\dfrac { 2\sqrt { 2 } { \mu  } _{ 0 }I }{ \pi  } \dfrac { { r }^{ 2 } }{ a } $------ (1)
    Mutual induction, M= $\dfrac { \phi  }{ { I } _{ 1 } } $

From (1), M= $\dfrac { 2\sqrt { 2 } { \mu  } _{ 0 } }{ \pi  } \dfrac { { r }^{ 2 } }{ a } $

Hence, $M\alpha \dfrac { { r }^{ 2 } }{ a } $

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

Which of the following statement is correct?

  1. when the magnetic flux linked with conducting loop is zero then emf induced is always zero

  2. when the emf induced in conducting loop is zero, then the magnetic flux linked with the loop must be zero

  3. transformer works on mutual induction

  4. all of these
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

 Statement is.

A) When the magnetic flux linked with conducting loop is zero then emf induced is always zero.
     $emf=\dfrac{d\phi}{dt}$
  If $\phi=0$, $emf=\dfrac{d0}{dt}=0$
B) when the emf induced in conducting loop is zero, then the magnetic flux linked with the loop must be zero.
    $emf=\dfrac{d\phi}{dt}=0$
    $d\phi=0$
   $\phi=constant$ magnetic flux is constant.
This is the wrong statement
C) The transformer works on mutual induction.
The correct statement is (A) and (C).


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

An electron originates at a point $A$ lying on the axis of a straight solenoid and moves with velocity $v$ at an angle $\alpha$ to the axis. The magnetic induction of the field is equal to $BA$ screen is oriented at right angles to the axis and is located at a distance $1$ from the point $a$. Find the distance from the axis to the point on the screen into which the electron strikes.

  1. $d = 5r\sin \left (\dfrac {\theta}{2}\right )$, Here $r = 2\dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$.

  2. $d = 2r\sin \left (\dfrac {\theta}{2}\right )$, Here $r = \dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$.

  3. $d = 3r\sin \left (\dfrac {\theta}{2}\right )$, Here $r = 3\dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$.

  4. $d = 4r\sin \left (\dfrac {\theta}{2}\right )$, Here $r = \dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The electron moves in a helical path in a uniform magnetic field. The radius of the helix is r = (mv*sin(alpha))/(eB) and the pitch angle/period determines the displacement. The distance from the axis is calculated using the geometry of the circular projection of the helical motion.