Miscellaneous (Mixed concepts) problems

Q 1 :

A conducting square loop initially lies in the XZ plane with its lower edge hinged along the X-axis. Only in the region $y \geq 0$ , there is a time dependent magnetic field pointing along the Z-direction, $\vec{B} (t) = B_{0} (\cos ω t) \hat{k},$

where $B_{0}$ is a constant. The magnetic field is zero everywhere else. At time $t = 0$ , the loop starts rotating with constant angular speed $ω$ about the X axis in the clockwise direction as viewed from the +X axis (as shown in the figure). Ignoring self-inductance of the loop and gravity, which of the following plots correctly represents the induced e.m.f. (V) in the loop as a function of time? [2025]

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(1)

Magnetic flux

$ϕ = B A \cos (90 ° - ω t) = (B_{0} \cos ω t) . A . \sin ω t$

$= \frac{B_{0} A}{2} \sin 2 ω t$

Induced emf, $e = - \frac{d ϕ}{d t}$

$= - \frac{B_{0} A}{2} (2 ω) \cos 2 ω t \propto - \cos 2 ω t,$ $ω_{eq} = 2 ω$

so, $T_{eq} = \frac{2 π}{ω_{eq}} = \frac{2 π}{2 ω} = \frac{π}{ω}$

Therefore, during the half revolution, there will be one cycle of emf. During the second half revolution, the loop will be out of the magnetic field, so the emf will remain zero.

Q 2 :

An infinitely long cylinder is kept parallel to an uniform magnetic field B directed along the positive z-axis. The direction of induced current as seen from the z-axis will be [2005]

zero
anticlockwise of the +ve z axis
clockwise of the +ve z axis
along the magnetic field

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(1)

As the cylinder is kept parallel to an uniform magnetic field, there is no change in magnetic flux and hence the induced current will be zero.

Q 3 :

A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be [2002]

halved
the same
doubled
quadrupled

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(2)

$Power, P = \frac{E^{2}}{R} = \frac{π r^{2}}{ρ ℓ} {(\frac{d ϕ}{d t})}^{2} = \frac{π r^{2}}{ρ ℓ} {[\frac{d}{d t} (N B A)]}^{2}$

$[Here, E = induced emf and R = ρ \frac{ℓ}{A}]$

or, $P = \frac{π r^{2}}{ρ ℓ} N^{2} A^{2} {(\frac{d B}{d t})}^{2}$ $\Rightarrow P \propto \frac{N^{2} r^{2}}{ℓ}$

When the number of turns is quadrupled and the wire radius is halved power,

$ρ' \propto \frac{{(4 N)}^{2} {(r / 2)}^{2}}{4 ℓ}$

$∴$ $\frac{P}{ρ'} = \frac{1}{1}$

$∴$ $Power remains the same.$

Q 4 :

As shown in the figure, P and Q are two coaxial conducting loops separated by some distance. When the switch S is closed, a clockwise current $I_{P}$ flows in P (as seen by E) and an induced current $I_{Q 1}$ flows in Q. The switch remains closed for a long time. When S is opened, a current $I_{Q 2}$ flows in Q. Then the direction of $I_{Q 1}$ and $I_{Q 2}$ (as seen by E) are [2002]

respectively clockwise and anti-clockwise
both clockwise
both anti-clockwise
respectively anti-clockwise and clockwise

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(4)

When switch $S$ is closed, a magnetic field is set up in the space around $P$ . The field lines threading $Q$ increase in the direction from right to left. According to Lenz's law, $I_{Q_{1}}$ will flow so as to oppose the cause or change and flow in the anticlockwise direction as seen by $E$ . Opposite is the case when $S$ is opened. $I_{Q_{2}}$ will be clockwise.

Q 5 :

A coil of wire having inductance and resistance has a conducting ring placed coaxially within it. The coil is connected to a battery at time $t = 0$ , so that a time-dependent current $I_{1} (t)$ starts flowing through the coil. If $I_{2} (t)$ is the current induced in the ring, and $B (t)$ is the magnetic field at the axis of the coil due to $I_{1} (t)$ , then as a function of time $(t > 0)$ , the product $I_{2} (t) B (t)$ [2000]

increases with time
decreases with time
does not vary with time
passes through a maximum

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(2)

A conducting ring is placed coaxially within a coil and the coil has inductance as well as resistance. The magnetic field at the centre of the coil $B (t) = μ_{0} n I_{1}$ .

As the current increases, $B$ will also increase with time till it reaches a maximum value (when the current becomes steady).

Induced emf in the ring

$e = - \frac{d ϕ}{d t} = - \frac{d}{d t} (\vec{B} \cdot \vec{A}) = - A \frac{d}{d t} (μ_{0} n I_{1})$

$∴$ Induced current in the ring

$I_{2} (t) = \frac{| e |}{R} = \frac{μ_{0} n A}{R} \frac{d I_{1}}{d t}$

[ $\frac{d I_{1}}{d t}$ decreases with time and hence $I_{2}$ also decreases with time.]

where $I_{1} = I_{\max} (1 - e^{- t / τ})$

The relevant graphs are as follows.

Hence, as a function of time $(t > 0)$ , the product $I_{2} (t) B (t)$ decreases with time.

Q 6 :

A uniform but time-varying magnetic field $B (t)$ exists in a circular region of radius $a$ and is directed into the plane of the paper, as shown. The magnitude of the induced electric field at point $P$ at a distance $r$ from the centre of the circular region [2000]

is zero
decreases as $1 / r$
increases as $r$
decreases as $1 / r^{2}$

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(2)

$\oint \vec{E} \cdot d \vec{ℓ} = \frac{d ϕ}{d t} = \frac{d}{d t} (\vec{B} \cdot \vec{A}) = \frac{d}{d t} (B A \cos 0 °) = π a^{2} \frac{d B}{d t}$

$\Rightarrow E (2 π r) = π a^{2} \frac{d B}{d t}$ For $r \geq a$ ,

$\Rightarrow E = \frac{a^{2}}{2 r} \frac{d B}{d t}$ $\Rightarrow E \propto \frac{1}{r}$

Hence, the magnitude of the induced electric field at a distance $r$ from the centre of the circular region decreases as $\frac{1}{r} .$

Q 7 :

A long straight wire carries a current, $I = 2$ ampere. A semi-circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both the rails are parallel to the wire. The wire, the rod and the rails lie in the same horizontal plane, as shown in the figure. Two ends of the semi-circular rod are at distances 1 cm and 4 cm from the wire. At time $t = 0$ , the rod starts moving on the rails with a speed $v = 3.0 m / s$ (see the figure).

A resistor $R = 1.4 Ω$ and a capacitor $C_{0} = 5.0 μ F$ are connected in series between the rails. At time $t = 0$ , $C_{0}$ is uncharged. Which of the following statement(s) is(are) correct?

$[μ_{0} = 4 π \times 10^{- 7} SI units, Take \ln 2 = 0.7]$ [2021]

Maximum current through $R$ is $1.2 \times 10^{- 6}$ ampere
Maximum current through $R$ is $3.8 \times 10^{- 6}$ ampere
Maximum charge on capacitor $C_{0}$ is $8.4 \times 10^{- 12}$ coulomb
Maximum charge on capacitor $C_{0}$ is $2.4 \times 10^{- 12}$ coulomb

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Select one or more options

(1, 3)

EMF developed across the semi-circular rod,

$ε = \int_{1}^{4} \frac{μ_{0} i}{2 π r} v d r = \frac{μ_{0} i v}{2 π} \ln (\frac{b}{a}) = \frac{μ_{0} i v}{2 π} \ln (\frac{4}{1}) = \frac{μ_{0} i v}{π} \ln 2$

$∴$ $ε = \frac{4 π \times 10^{- 7} \times 2 \times 3 \times 0.7}{π} = 24 \times 7 \times 10^{- 8} = 1.68 \times 10^{- 6} V$

Therefore, maximum current through $R$ ,

$i_{\max} = \frac{ε}{R} = \frac{1.68 \times 10^{- 6}}{1.4} = 1.2 \times 10^{- 6} A$

And maximum charge on capacitor $C_{0}$ ,

$Q_{\max} = C_{0} E = 5 \times 10^{- 6} \times 1.68 \times 10^{- 6} = 8.4 \times 10^{- 12} C$

Q 8 :

A source of constant voltage $V$ is connected to a resistance $R$ and two ideal inductors $L_{1}$ and $L_{2}$ through a switch $S$ as shown. There is no mutual inductance between the two inductors. The switch $S$ is initially open. At $t = 0$ , the switch is closed and current begins to flow. Which of the following options is/are correct? [2017]

After a long time, the current through $L_{1}$ will be $\frac{V}{R} \frac{L_{2}}{L_{1} + L_{2}}$
After a long time, the current through $L_{2}$ will be $\frac{V}{R} \frac{L_{1}}{L_{1} + L_{2}}$
The ratio of the currents through $L_{1}$ and $L_{2}$ is fixed at all times $(t > 0)$
At $t = 0$ , the current through the resistance $R$ is $\frac{V}{R}$

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Select one or more options

(1, 2, 3)

At $t = 0$ inductors $L_{1}$ and $L_{2}$ will offer infinite resistance, hence current through circuit is zero.

After a long time, the current through the resistor is constant.

I will divide into two parts through $L_{1}$ and $L_{2}$ , which are in parallel.

$∴$ $I_{1} L_{1} = I_{2} L_{2}$ $(I = I_{1} + I_{2})$

$I_{1} = \frac{V}{R} (\frac{L_{2}}{L_{1} + L_{2}})$ and $I_{2} = \frac{V}{R} (\frac{L_{1}}{L_{1} + L_{2}})$

Also, the ratio of currents through $L_{1}$ and $L_{2}$ is fixed at all times.

At $t = 0,$ $I \approx 0$

Q 9 :

A circular insulated copper wire loop is twisted to form two loops of area A and 2A as shown in the figure. At the point of crossing, the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field $\vec{B}$ points into the plane of the paper. At $t = 0$ , the loop starts rotating about the common diameter as axis with a constant angular velocity $ω$ in the magnetic field. Which of the following options is/are correct? [2017]

The emf induced in the loop is proportional to the sum of the areas of the two loops.
The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of maximum emf induced in the smaller loop alone.
The net emf induced due to both the loops is proportional to $\cos ω t .$
The rate of change of the flux is maximum when the plane of the loop is perpendicular to the plane of the paper.

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Select one or more options

(2, 4)

The net magnetic flux through the loops at time $t$

$ϕ = B (2 A) \cos ω t - B A \cos ω t = B (2 A - A) \cos ω t = B A \cos ω t$

$∴$ $| \frac{d ϕ}{d t} |$ $= B ω A \sin ω t$

So, $| \frac{d ϕ}{d t} |$ is maximum when $ω t = \frac{π}{2}$

The emf induced in the smaller loop is

$ε_{smaller} = - \frac{d}{d t} (B A \cos ω t) = B ω A \sin ω t$

$∴$ Amplitude of maximum net emf induced in both the loops = Amplitude of maximum emf induced in the smaller loop alone.

Q 10 :

At time $t = 0$ , terminal A in the circuit shown in the figure is connected to B by a key and an alternating current $I (t) = I_{0} \cos (ω t),$ with $I_{0} = 1 A$ and $ω = 500 {rad s}^{- 1}$ starts flowing in it with the initial direction shown in the figure. At $t = \frac{7 π}{6 ω},$ the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If $C = 20 μ F,$ $R = 10 Ω$ and the battery is ideal with emf of 50 V, identify the correct statement(s). [2014]

Magnitude of the maximum charge on the capacitor before $t = \frac{7 π}{6 ω}$ is $1 \times 10^{- 3} C$
The current in the left part of the circuit just before $t = \frac{7 π}{6 ω}$ is clockwise.
Immediately after A is connected to D, the current in R is 10 A
$Q = 2 \times 10^{- 3} C$

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Select one or more options

(3, 4)

For maximum charge on the capacitor, $\frac{d Q}{d t} = I = 0$

$I = I_{0} \cos ω t = \cos 500 t$

Till $t = \frac{7 π}{6 ω}$ , the charge will be maximum at $t = \frac{π}{2 ω}$

$Q' = \int_{0}^{π / (2 ω)} \cos 500 t d t = {[\frac{\sin 500 t}{500}]}_{0}^{π / 2 ω}$

$= \frac{1}{500} \sin (500 \times \frac{π}{2 \times 500}) = \frac{1}{500} C$

i.e., $Q_{\max} = \frac{1}{500} C = 2 \times 10^{- 3} C$

From the graph it is clear that just before $t = \frac{7 π}{6 ω},$ the current is in the anticlockwise direction. Immediately after A is connected to D.

At $t = \frac{7 π}{6 ω},$ the charge on the upper plate of the capacitor is

$\int_{0}^{7 π / (6 ω)} \cos 500 t d t = \frac{1}{500} \sin (500 \times \frac{7 π}{6 \times 500})$

$= - \frac{1}{500} \times \frac{1}{2} = - 10^{- 3} C$

Now applying KVL,

$50 + \frac{10^{- 3}}{20 \times 10^{- 6}} - 10 i = 0$ $\Rightarrow i = 10 A$

The maximum charge on $C$ , $Q = C V = 20 \times 10^{- 6} \times 50 = 10^{- 3} C$

Therefore, the total charge flown from the battery is $2 \times 10^{- 3} C$

Q 11 :

A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity $ω$ . This can be considered as equivalent to a loop carrying a steady current $\frac{Q ω}{2 π} .$ A uniform magnetic field along the positive $z$ -axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant $g$ . [2013]

Q. The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is

$\frac{B R}{4}$
$\frac{B R}{2}$
$B R$
$2 B R$

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(2)

$\int \vec{E} \cdot d \vec{ℓ} = - \frac{d ϕ}{d t} = - \frac{d}{d t} (B π R^{2}) = - π R^{2} \frac{d B}{d t}$

Given, $\frac{d B}{d t} = B$

therefore, $\int \vec{E} \cdot d \vec{ℓ} = - π R^{2} B$

$∴$ $E (2 π R) = - π R^{2} B$

$∴$ $E = - \frac{B R}{2}$

Q 12 :

A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity $ω$ . This can be considered as equivalent to a loop carrying a steady current $\frac{Q ω}{2 π} .$ A uniform magnetic field along the positive $z$ -axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant $g$ . [2013]

Q. The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is

$- γ B Q R^{2}$
$- γ \frac{B Q R^{2}}{2}$
$γ \frac{B Q R^{2}}{2}$
$γ B Q R^{2}$

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(2)

Given $M = γ L$

$∴$ $M = γ m ω R^{2}$

$∴$ $Δ M = γ m (Δ ω) R^{2}$ $\dots (1)$

But $Δ ω = \frac{Q B}{2 m}$ $\dots (2)$

From Eqs. (1) and (2),

$Δ M = - γ m (\frac{Q B}{2 m}) R^{2} = - \frac{γ B Q R^{2}}{2}$

Here, the negative sign shows that the change is opposite to the direction of the magnetic field B.

Hence, (2) is the correct option.

Q 13 :

Statement-1 : A vertical iron rod has coil of wire wound over it at the bottom end. An alternating current flows in the coil. The rod goes through a conducting ring as shown in the figure. The ring can float at a certain height above the coil.

Statement-2 : In the above situation, a current is induced in the ring which interacts with the horizontal component of the magnetic field to produce an average force in the upward direction. [2007]

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
Statement-1 is True, Statement-2 is False
Statement-1 is False, Statement-2 is True

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(1)

The horizontal component of the magnetic field $B_{H}$ interacts with the induced current produced in the conducting ring which produces an average force in the upward direction. This is in accordance with Fleming's left hand rule.

Topic Question Set

Q 2 :

An infinitely long cylinder is kept parallel to an uniform magnetic field B directed along the positive z-axis. The direction of induced current as seen from the z-axis will be [2005]

Q 3 :

A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be [2002]

Q 6 :

A uniform but time-varying magnetic field $B (t)$ exists in a circular region of radius $a$ and is directed into the plane of the paper, as shown. The magnitude of the induced electric field at point $P$ at a distance $r$ from the centre of the circular region [2000]

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Topic Question Set

Q 2 : An infinitely long cylinder is kept parallel to an uniform magnetic field B directed along the positive z-axis. The direction of induced current as seen from the z-axis will be [2005]

Q 3 : A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be [2002]

Q 6 : A uniform but time-varying magnetic field B(t) exists in a circular region of radius a and is directed into the plane of the paper, as shown. The magnitude of the induced electric field at point P at a distance r from the centre of the circular region [2000]

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Q 2 :

An infinitely long cylinder is kept parallel to an uniform magnetic field B directed along the positive z-axis. The direction of induced current as seen from the z-axis will be [2005]

Q 3 :

A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be [2002]

Q 6 :

A uniform but time-varying magnetic field $B (t)$ exists in a circular region of radius $a$ and is directed into the plane of the paper, as shown. The magnitude of the induced electric field at point $P$ at a distance $r$ from the centre of the circular region [2000]