Electromagnetic Induction

JEE Main level, single correct, medium

50 of 153

Question 50 of 153

A rectangular loop of length 2.5m2.5m and width 2m2m is placed at 60∘60^{\circ} to a magnetic field of 4T4T. The loop is removed from the field in 10sec⁡10\sec. The average emf induced in the loop during this time is

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Answer: (C) +1V+1V

The formula to calculate the change in the magnetic flux associated with the given loop can be written as
∆ϕ=−BAcos⁡θ=−B(lb)cos⁡θ...(1)\begin{matrix}∆\phi & = & -BA\cos \theta \\ & = & -B\left(lb\right)\cos \theta ...\left(1\right)\end{matrix}
The formula to calculate the induced emf is given by
ϵ=−∆ϕ∆t=B(lb)cos⁡θ∆t...(2)\begin{matrix}\epsilon & = & -\dfrac{∆\phi }{∆t} \\ & = & \dfrac{B\left(lb\right)\cos \theta }{∆t}...\left(2\right)\end{matrix}
From equation (2), it follows that
ϵ=4×(2.5×2)×cos⁡60∘10V=1V\begin{matrix}\epsilon & = & \dfrac{4\times \left(2.5\times 2\right)\times \cos 60^{\circ}}{10}V \\ & = & 1V\end{matrix}

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