Electromagnetic Induction

JEE Main level, single correct, medium

3 of 153

Question 3 of 153

A square loop of side 22 cm is placed in a time varying magnetic field with magnitude as B=0.4sin⁡(300t)B = 0.4\sin(300t) Tesla. The normal to the plane of loop makes an angle of 60°60° with the field. The maximum induced emf produced in the loop is __________ mV.

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Answer: (D) 2424

Given:
Side of the square loop, a=2a = 2 cm =2×10−2= 2 \times 10^{-2} m
Area of the loop, A=a2=4×10−4A = a^2 = 4 \times 10^{-4} m2^2
Magnetic field, B=0.4sin⁡(300t)B = 0.4 \sin(300t) T
Angle between the normal to the plane and the magnetic field, θ=60∘\theta = 60^\circ

The magnetic flux through the loop is given by:
Φ=B⃗⋅A⃗=BAcos⁡θ\Phi = \vec{B} \cdot \vec{A} = B A \cos\theta
Φ=(0.4sin⁡(300t))(4×10−4)cos⁡(60∘)\Phi = (0.4 \sin(300t))(4 \times 10^{-4}) \cos(60^\circ)
Φ=1.6×10−4sin⁡(300t)×12\Phi = 1.6 \times 10^{-4} \sin(300t) \times \dfrac{1}{2}
Φ=0.8×10−4sin⁡(300t)\Phi = 0.8 \times 10^{-4} \sin(300t) Wb

By Faraday's law of induction, the induced emf is:
e=−dΦdte = -\dfrac{d\Phi}{dt}
e=−ddt(0.8×10−4sin⁡(300t))e = -\dfrac{d}{dt} (0.8 \times 10^{-4} \sin(300t))
e=−0.8×10−4×300cos⁡(300t)e = -0.8 \times 10^{-4} \times 300 \cos(300t)
e=−240×10−4cos⁡(300t)e = -240 \times 10^{-4} \cos(300t) V
e=−24×10−3cos⁡(300t)e = -24 \times 10^{-3} \cos(300t) V

The maximum induced emf is the amplitude of ee:
emax=24×10−3e_{\text{max}} = 24 \times 10^{-3} V =24= 24 mV

Answer: 2424

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