Electromagnetic Induction

JEE Main level, numerical answer, medium

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Question 4 of 153

A circular loop of radius 20 cm20\text{ cm} and resistance 2 Ω2\ \Omega is placed in a time varying magnetic field B⃗=(2t2+2t+3) T\vec{B} = (2t^2 + 2t + 3)\text{ T}. At t=0t = 0, for the plane of the loop being perpendicular to the magnetic field and, the induced current in the loop at t=3 st = 3\text{ s} is α50 A\dfrac{\alpha}{50}\text{ A}. The value of α\alpha is _______.
(Take π=22/7\pi = 22/7)

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Answer: 44

The magnetic flux linked with the circular loop is given by Φ=BAcos⁡θ\Phi = BA \cos \theta. Since the plane of the loop is perpendicular to the magnetic field, the angle between the area vector and the magnetic field is θ=0∘\theta = 0^{\circ}.

Φ=B⋅πr2\Phi = B \cdot \pi r^2

Given r=20 cm=0.2 mr = 20\text{ cm} = 0.2\text{ m}, the area of the loop is:

A=π(0.2)2=0.04π m2A = \pi (0.2)^2 = 0.04\pi\text{ m}^2

The induced emf in the loop is given by Faraday's law of induction:

∣E∣=dΦdt=AdBdt|E| = \dfrac{d\Phi}{dt} = A \dfrac{dB}{dt}

Given the magnetic field B=2t2+2t+3B = 2t^2 + 2t + 3, we differentiate it with respect to time tt:

dBdt=ddt(2t2+2t+3)=4t+2\dfrac{dB}{dt} = \dfrac{d}{dt}(2t^2 + 2t + 3) = 4t + 2

At t=3 st = 3\text{ s}, the rate of change of the magnetic field is:

dBdt=4(3)+2=14 T/s\dfrac{dB}{dt} = 4(3) + 2 = 14\text{ T/s}

Now, substituting the values into the emf equation:

∣E∣=0.04π×14=0.56π V|E| = 0.04\pi \times 14 = 0.56\pi\text{ V}

The induced current II in the loop is:

I=∣E∣R=0.56π2=0.28π AI = \dfrac{|E|}{R} = \dfrac{0.56\pi}{2} = 0.28\pi\text{ A}

Using π=227\pi = \dfrac{22}{7}, we get:

I=0.28×227=0.04×22=0.88 AI = 0.28 \times \dfrac{22}{7} = 0.04 \times 22 = 0.88\text{ A}

We are given that the induced current is α50 A\dfrac{\alpha}{50}\text{ A}. Equating the two values:

α50=0.88\dfrac{\alpha}{50} = 0.88

α=0.88×50=44\alpha = 0.88 \times 50 = 44

Answer: 4444

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