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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2014
  • Medium
  • Numerical

A coil of circular cross-section having \(1000\, turns\) and \(4\, cm^2\) face area is placed with its axis parallel to a magnetic field which decreases by \(10^{-2}\, Wb\, m^{-2}\) in \(0.01\, s\). The \(e.m.f.\) induced in the coil is....\(mV\)

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Answer & Solution

Correct Answer

(A) \(400\)

Step-by-step Solution

Detailed explanation

Given: \(Na\) of turns \(N=1000\)

Face area, \(A=4 \,\mathrm{cm}^{2}=4 \times 10^{-4} \,\mathrm{m}^{2}\)

Change in magnetic field,

\(\Delta B=10^{-2} \,\mathrm{wbm}^{-2}\)

Time taken, \(t=0.01\, \mathrm{s}=10^{-2}\, \mathrm{sec}\)

\(Emf\) induced in the coil \(e=?\)

Applying formula,

Induced \(emf\), \(e=\frac{-d \phi}{d t}\)

\(=N\left(\frac{\Delta B}{\Delta t}\right) A \cos \theta\)

\(=\frac{1000 \times 10^{-2} \times 4 \times 10^{-4}}{10^{-2}}=400\, \mathrm{mV}\)