MHT CET · Physics · Electromagnetic Induction
A magnetic field of \(2 \times 10^{-2} \mathrm{~T}\) acts at right angles to a coil of area \(100 \mathrm{~cm}^2\) with 50 turns. The average e.m.f. induced in the coil is 0.1 V , when it is removed from the field in time \(t\). The value of ' \(t\) ' is (in second)
- A 0.1 s
- B \(\quad 0.01 \mathrm{~s}\)
- C 1 s
- D 20 s
Answer & Solution
Correct Answer
(A) 0.1 s
Step-by-step Solution
Detailed explanation
Using Lenz law,
Induced e.m.f. is given by, \(e=-N \frac{d \phi}{d t}\)
Change in magnetic flux \(=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{dB}}{\mathrm{dt}} \mathrm{A} \cos \theta\)
From (i),
\(\begin{aligned} e & =-N \frac{d B}{d t} A \cos \theta \\ \therefore \quad e & =-\frac{N\left(B_2-B_1\right) A \cos \theta}{t}\end{aligned}\)
\(\begin{array}{ll}\therefore & t=\frac{-(50) \times\left(0-2 \times 10^{-2}\right) \times\left(100 \times 10^{-4}\right) \times \cos 0^{\circ}}{0.1} \\ \therefore & t=0.1 \mathrm{~s}\end{array}\)
Induced e.m.f. is given by, \(e=-N \frac{d \phi}{d t}\)
Change in magnetic flux \(=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{dB}}{\mathrm{dt}} \mathrm{A} \cos \theta\)
From (i),
\(\begin{aligned} e & =-N \frac{d B}{d t} A \cos \theta \\ \therefore \quad e & =-\frac{N\left(B_2-B_1\right) A \cos \theta}{t}\end{aligned}\)
\(\begin{array}{ll}\therefore & t=\frac{-(50) \times\left(0-2 \times 10^{-2}\right) \times\left(100 \times 10^{-4}\right) \times \cos 0^{\circ}}{0.1} \\ \therefore & t=0.1 \mathrm{~s}\end{array}\)
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