KCET · Physics · Electromagnetic Induction
A magnetic field of flux denisty \(1.0 \mathrm{~Wb} \mathrm{~m}^{-2}\) acts normal to a 80 turn coil of \(0.01 \mathrm{~m}^2\) area. If this coil is removed from the field in \(0.2 \mathrm{~s}\), then the emf induced in it is
- A \(8 \mathrm{~V}\)
- B \(0.8 \mathrm{~V}\)
- C \(5 \mathrm{~V}\)
- D \(4 \mathrm{~V}\)
Answer & Solution
Correct Answer
(D) \(4 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Given, magnetic flux density,
\(B=1 \mathrm{~Wb} / \mathrm{m}^2\)
Number of turns, \(N=80\)
Area of coil, \(A=0.01 \mathrm{~m}^2\)
\(\Delta t=0.2 \mathrm{~s}\)
According to Faraday's law of electromagnetic induction, induced emf
\(\begin{aligned}|e| & =\frac{N \Delta \phi}{\Delta t}=N \cdot \frac{B A}{\Delta t} \quad[\because \Delta \phi=B A] \\& =\frac{80 \times 1 \times 0.01}{0.2}=4 \mathrm{~V}\end{aligned}\)
\(B=1 \mathrm{~Wb} / \mathrm{m}^2\)
Number of turns, \(N=80\)
Area of coil, \(A=0.01 \mathrm{~m}^2\)
\(\Delta t=0.2 \mathrm{~s}\)
According to Faraday's law of electromagnetic induction, induced emf
\(\begin{aligned}|e| & =\frac{N \Delta \phi}{\Delta t}=N \cdot \frac{B A}{\Delta t} \quad[\because \Delta \phi=B A] \\& =\frac{80 \times 1 \times 0.01}{0.2}=4 \mathrm{~V}\end{aligned}\)
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