A conducting loop in the shape of a right angled isosceles triangle of height 10 cm is kept such that the 90 ^\circ vertex is very close to an infinitely long conducting wire (see the figure). The wire is electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The current in the triangular loop is in counterclockwise direction and increased at a constant rate of 10 As - 1. Which of the following statement(s) is (are) true?

There is a repulsive force between the wire and the loop., The magnitude of induced emf in the wire is ( \mu _0 \over \pi ) volt

There is a repulsive force between the wire and the loop.

If the loop is rotated at a constant angular speed about the wire, an additional emf of ( \mu _0 \over \pi ) volt is induced in the wire

The magnitude of induced emf in the wire is ( \mu _0 \over \pi ) volt

The induced current in the wire is in opposite direction to the current along the hypotenuse.

Electromagnetic Induction PYQ — JEE Advanced 2016 · Shift 1 · Q46 with Solution

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Correct answer: A, C

Geometry of the loop

The loop is a right-angled isosceles triangle of height $10\,\text{cm}=0.1\,\text{m}$ .

If the right-angle vertex is nearest to the long straight wire and the hypotenuse is parallel to the wire, then the two equal perpendicular sides are at $45^\circ$ to the wire.

Let the wire be along the $z$ -direction, and let the perpendicular distance from the wire to a strip of the triangle be $x$ .

At distance $x$ from the vertex, the length of the strip parallel to the wire is

\ell(x)=2x

for $0\le x\le h$ , where $h=0.1\,\text{m}$ .

This is because the two sides make $45^\circ$ with the hypotenuse/wire.

Magnetic field due to current in the triangular loop at the wire

We need the flux linkage with the long wire due to the current in the triangular loop. A convenient way is to calculate the mutual inductance.

Magnetic field due to a long straight wire at distance $x$ is

B=\frac{\mu_0 I}{2\pi x}.

So the flux through an elemental strip of width $dx$ and length $\ell(x)=2x$ is

d\Phi = B\,dA = \frac{\mu_0 I}{2\pi x}(2x\,dx)=\frac{\mu_0 I}{\pi}dx.

Integrating from $x=0$ to $x=h$ ,

\Phi = \int_0^h \frac{\mu_0 I}{\pi}dx =\frac{\mu_0 I h}{\pi}.

Hence the mutual inductance is

M=\frac{\Phi}{I}=\frac{\mu_0 h}{\pi}.

With $h=0.1\,\text{m}$ ,

M=\frac{\mu_0}{10\pi}.

Induced emf in the wire due to changing current in the loop

Given

\frac{dI}{dt}=10\,\text{A s}^{-1}.

Therefore induced emf magnitude in the wire is

\mathcal E = M\frac{dI}{dt} =\frac{\mu_0}{10\pi}\times 10 =\frac{\mu_0}{\pi}\,\text{V}.

So option C is correct.

Direction of induced current in the wire

The current in the triangular loop is counterclockwise. Along the hypotenuse (which is parallel to the wire), the current flows upward.

By the right-hand rule, the magnetic field produced by this loop at the wire is into/out of the page in a definite sense; since the loop current is increasing, the magnetic flux through the wire due to the loop increases.

By Lenz's law, the induced current in the wire must oppose this increase.

This makes the current in the wire flow in the same direction as the current in the hypotenuse, so that the force between the wire and the hypotenuse is repulsive-dominant overall (nearest parallel segment dominates), giving repulsion between wire and loop.

Thus:

Option D is false.
Option A is correct.

Check option B: rotation about the wire

If the loop is rotated about the wire at constant angular speed, the shape and relative radial distribution of area about the wire do not change in a way that changes the mutual inductance with the wire. The flux linkage depends on the geometry relative to the wire, and under such rotation about the wire, no new constant additional emf of

\frac{\mu_0}{\pi}\,\text{V}

is generated.

So option B is false.

Final evaluation of options

A: True
B: False
C: True
D: False

Thus the correct statements are

\boxed{A,\ C}

Comparison with stored answer

Stored correct answer: A, C

This matches the derived answer.

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View written solutionFree

Correct answer: A, C

Geometry of the loop

The loop is a right-angled isosceles triangle of height $10\,\text{cm}=0.1\,\text{m}$ .

If the right-angle vertex is nearest to the long straight wire and the hypotenuse is parallel to the wire, then the two equal perpendicular sides are at $45^\circ$ to the wire.

Let the wire be along the $z$ -direction, and let the perpendicular distance from the wire to a strip of the triangle be $x$ .

At distance $x$ from the vertex, the length of the strip parallel to the wire is

\ell(x)=2x

for $0\le x\le h$ , where $h=0.1\,\text{m}$ .

This is because the two sides make $45^\circ$ with the hypotenuse/wire.

Magnetic field due to current in the triangular loop at the wire

We need the flux linkage with the long wire due to the current in the triangular loop. A convenient way is to calculate the mutual inductance.

Magnetic field due to a long straight wire at distance $x$ is

B=\frac{\mu_0 I}{2\pi x}.

So the flux through an elemental strip of width $dx$ and length $\ell(x)=2x$ is

d\Phi = B\,dA = \frac{\mu_0 I}{2\pi x}(2x\,dx)=\frac{\mu_0 I}{\pi}dx.

Integrating from $x=0$ to $x=h$ ,

\Phi = \int_0^h \frac{\mu_0 I}{\pi}dx =\frac{\mu_0 I h}{\pi}.

Hence the mutual inductance is

M=\frac{\Phi}{I}=\frac{\mu_0 h}{\pi}.

With $h=0.1\,\text{m}$ ,

M=\frac{\mu_0}{10\pi}.

Induced emf in the wire due to changing current in the loop

Given

\frac{dI}{dt}=10\,\text{A s}^{-1}.

Therefore induced emf magnitude in the wire is

\mathcal E = M\frac{dI}{dt} =\frac{\mu_0}{10\pi}\times 10 =\frac{\mu_0}{\pi}\,\text{V}.

So option C is correct.

Direction of induced current in the wire

The current in the triangular loop is counterclockwise. Along the hypotenuse (which is parallel to the wire), the current flows upward.

By Lenz's law, the induced current in the wire must oppose this increase.

Thus:

Option D is false.
Option A is correct.

Check option B: rotation about the wire

\frac{\mu_0}{\pi}\,\text{V}

is generated.

So option B is false.

Final evaluation of options

A: True
B: False
C: True
D: False

Thus the correct statements are

\boxed{A,\ C}

Comparison with stored answer

Stored correct answer: A, C

This matches the derived answer.

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