A rectangular loop of length 2.5 m and width 2 m is placed at 60^ to a magnetic field of 4 T. The loop is removed from the field in 10 s. The average emf induced in the loop during this time is:

Given: length of loop = 2.5 m, width = 2 m, magnetic field = 4 T, angle = 60^, time taken = 10 s. Find: the average induced emf. Using Faraday's law of electromagnetic induction, emf = - t where is the change in magnetic flux. The initial magnetic flux through the loop is i = B A The area of the loop is A = length width = 2.5 2 = 5 m^2 So, i = 4 5 60^ = 4 5 0.5 = 10 Wb When the loop is removed from the magnetic field, the final magnetic flux becomes f = 0 Therefore, the change in flux is = f - i = 0 - 10 = -10 Wb Substitute directly into Faraday's law: Average emf = - t = - 0 - (4 (2.5 2) 60^)10 = 4 (2.5 2) 12 = 10 Wb Hence, Average emf = - -1010 = +1 V Therefore, the correct option is C.

A rectangular loop of length 2.5 m and width 2 m is placed at 60^… — Solution | JEE Physics

MCQEasyJEE 2024Faraday's Laws of EMI

A rectangular loop of length $2.5 \, \text{m}$ and width $2 \, \text{m}$ is placed at $60^\circ$ to a magnetic field of $4 \, \text{T}$ . The loop is removed from the field in $10 \, \text{s}$ . The average emf induced in the loop during this time is:

A
$-2 \, \text{V}$
B
$+2 \, \text{V}$
C
$+1 \, \text{V}$
D
$-1 \, \text{V}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: length of loop = $2.5 \, \text{m}$ , width = $2 \, \text{m}$ , magnetic field = $4 \, \text{T}$ , angle = $60^\circ$ , time taken = $10 \, \text{s}$ .

Find: the average induced emf.

Using Faraday's law of electromagnetic induction,

\text{emf} = -\frac{\Delta \Phi}{\Delta t}

where $\Delta \Phi$ is the change in magnetic flux.

The initial magnetic flux through the loop is

\Phi_i = B \cdot A \cdot \cos \theta

The area of the loop is

A = \text{length} \times \text{width} = 2.5 \times 2 = 5 \, \text{m}^2

So,

\Phi_i = 4 \times 5 \times \cos 60^\circ = 4 \times 5 \times 0.5 = 10 \, \text{Wb}

Flux Change Calculation

When the loop is removed from the magnetic field, the final magnetic flux becomes

\Phi_f = 0

Therefore, the change in flux is

\Delta \Phi = \Phi_f - \Phi_i = 0 - 10 = -10 \, \text{Wb}

Direct Substitution

Substitute directly into Faraday's law:

\text{Average emf} = -\frac{\Delta \Phi}{\Delta t} = -\frac{0 - \left(4 \times (2.5 \times 2) \cos 60^\circ\right)}{10}

\Delta \Phi = 4 \times (2.5 \times 2) \times \frac{1}{2} = 10 \, \text{Wb}

Hence,

\text{Average emf} = -\frac{-10}{10} = +1 \, \text{V}

Therefore, the correct option is C.

Common mistakes

Using $\sin 60^\circ$ instead of $\cos 60^\circ$ for magnetic flux. Flux is given by $\Phi = BA\cos\theta$ for the angle used in the solution, so the correct trigonometric factor must match the stated relation.
Ignoring the negative sign in Faraday's law. The sign indicates Lenz's law, and here it must be applied to the change in flux before determining the final numerical value of emf.
Forgetting that after removal from the field the final flux is $0$ . Once the loop is completely outside the magnetic field region, $\Phi_f = 0$ must be used in the flux-change calculation.

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