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....\(mV\)
Pick an option to check your answer.
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 \(emf\), \(e=\frac{-d \phi}{d t}\)
\(=N\left(\frac{\Delta B}{\Delta t}\right) A \cos \theta\)
\(=\frac{1000 \times 10^{-2} \times 4 \times 10^{-4}}{10^{-2}}=400\, \mathrm{mV}\)
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