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

Two conducting circular loops of radii $R _{1}$ and $R _{2}$ are placed in the same plane with their centres coinciding. If $R _{1} \gg R _{2}$, the mutual inductance $M$ between them will be directly proportional to

  1. $R _{1}/R _{2}$

  2. $R _{2}/R _{1}$

  3. $R _{1}^{2}/R _{2}$

  4. $R _{2}^{2}/R _{1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For two concentric loops where R1 >> R2, the magnetic field produced by the larger loop (R1) at its center is B = (mu0 * I) / (2 * R1). The flux through the smaller loop is Phi = B * Area2 = (mu0 * I * pi * R2^2) / (2 * R1). Since M = Phi / I, M = (mu0 * pi * R2^2) / (2 * R1), which is proportional to R2^2 / R1.

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

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

  1. increases when they are bought nearer.

  2. depends on the current passing through the coils.

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

  4. is not same as $M _{21}$ of coil 2 with respect to coil 1.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mutual inductance depends on the geometry, orientation, and separation of the coils. Bringing them closer increases the magnetic flux linkage between them, thereby increasing the mutual inductance.

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

A long solenoid  of diameter $0.1\ m$ has $2 \times {10^4}$ turns per metre.At the centre of the solenoid, a coil of $100$ turns and radius $0.01\ m$ is placed with its axis coinciding with the solenoid axis.The current in the solenoid reduces at a constant rate to $0\ A$ from $4\ A$ in $0.05\ s$. If the resistance of the coil is $10 \ {\pi ^2}\Omega ,$ the total charge flowing through the coil during this time is.

  1. $32\ \pi \mu C$

  2. $16\ \mu C$

  3. $32\ \mu C$

  4. $16\ \pi \mu C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,

Number of turns, $n=100$

Radius, $r=0.01\,m$

Resistance, $R=10\pi^2 \Omega$

As we know,

$\epsilon=-N\dfrac{d\phi}{dt}$

$=\dfrac{\epsilon}{R}=-\dfrac NR\dfrac{d\phi}{dt}$,   $\Delta I=-\dfrac NR\dfrac{d\phi}{dt}$

$\dfrac{\Delta}{\Delta t}=-\dfrac NR\dfrac{\Delta\phi}{\Delta t}\implies \Delta q=-[\dfrac NR(\dfrac{\Delta \phi}{\Delta t})]\Delta t$

$-$ve sign shoes that induced emf opposes the change in flux.

$\Delta q=\dfrac{\mu _0 ni\pi r^2}{R}$

$\Delta q=\dfrac{4\pi\times 10^{-7}\times 100\times 4\times \pi\times (0.01)^2}{10\pi^2}=32\mu C$
Multiple choice mutual inductance electromagnetic induction electromagnetic induction and alternating currents physics

Two coils A and B have mutual inductance $2\times { 10 }^{ -2 }$ henry. If the current in the primary is $i=5\sin { \left( 10\pi t \right)  } $ then the maximum value of e.m.f.induced in coil B is 

  1. $\pi \quad volt$

  2. $\pi /2volt$

  3. $\pi /3volt$

  4. $\pi /4volt$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The induced EMF is E = M * (di/dt). Given i = 5 * sin(10 * pi * t), di/dt = 5 * 10 * pi * cos(10 * pi * t) = 50 * pi * cos(10 * pi * t). The maximum EMF is E_max = M * (di/dt)_max = 2 * 10^-2 * 50 * pi = 1 * pi = pi V.

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

Two coils A and B have mutual inductance $2\times { 10 }^{ -2 }$ henry. If the current in the primary is $i=5\sin { \left( 10\pi t \right)  } $ then the maximum value of e.m.f. induced in coil B is

  1. $\pi \quad volt$

  2. $\pi /2volt$

  3. $\pi /3volt$

  4. $\pi /4volt$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a duplicate of 530241. The induced EMF is E = M * (di/dt). With i = 5 * sin(10 * pi * t), di/dt = 50 * pi * cos(10 * pi * t). E_max = 2 * 10^-2 * 50 * pi = pi V.

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

The electric field of an electromagnetic wave is given by, $E=(50N^{-1})\, \sin { \omega  } (t-x/c)$. Find the energy contained in a cylinder of cross section $10cm^2$ and length $50 cm$ along the x-axis.

  1. $5.5\times 10^{-12}J$

  2. $4.5\times 10^{-12}J$

  3. $5\times 10^{-13}J$

  4. $3.5\times 10^{-10}J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The energy density of an EM wave is u = (1/2) * epsilon0 * E^2 + (1/2) * (B^2 / mu0). For an EM wave, the average energy density is u_avg = (1/2) * epsilon0 * E0^2. The total energy is U = u_avg * Volume. Volume = Area * length = 10 * 10^-4 m^2 * 0.5 m = 5 * 10^-4 m^3. E0 = 50 V/m. u_avg = 0.5 * 8.85 * 10^-12 * 50^2 = 1.1 * 10^-8 J/m^3. U = 1.1 * 10^-8 * 5 * 10^-4 = 5.5 * 10^-12 J.