KCET · Physics · Electromagnetic Induction
A rectangular coil of 100 turns and size \(0.1 \mathrm{~m} \mathrm{} \times 0.05 \mathrm{~m}\) is placed perpendicular to a magnetic field of \(0.1 \mathrm{~T}\). If the field drops to \(0.05 \mathrm{~T}\) in \(0.05 \mathrm{~s}\), the magnitude of the e.m.f. iniduced in the coil is
- A \(\sqrt{2}\)
- B \(\sqrt{3}\)
- C \(\sqrt{0.6}\)
- D \(\sqrt{6}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{6}\)
Step-by-step Solution
Detailed explanation
Given, \(n=100\) turns, \(A=0.1 \times 0.05 \mathrm{~m}^{2}\),
\(B_{1}=0.1 \mathrm{~T}, \quad B_{2}=0.05 \mathrm{~T}, d t=0.05 \mathrm{~s}\)
We know that,
\(e=\left|\frac{-d \phi}{d t}\right|=\frac{d}{d t}(n B A \cos \theta)=\frac{n A d B \cos \theta}{d t}\)
Here, \(\theta=0^{\circ}\)
\(\therefore e=\frac{n A d B}{d t}=\frac{100 \times 0.1 \times 0.05 \times(0.1-0.05)}{0.05}\)
\(\Rightarrow e=0.5 \mathrm{~V}\)
\(B_{1}=0.1 \mathrm{~T}, \quad B_{2}=0.05 \mathrm{~T}, d t=0.05 \mathrm{~s}\)
We know that,
\(e=\left|\frac{-d \phi}{d t}\right|=\frac{d}{d t}(n B A \cos \theta)=\frac{n A d B \cos \theta}{d t}\)
Here, \(\theta=0^{\circ}\)
\(\therefore e=\frac{n A d B}{d t}=\frac{100 \times 0.1 \times 0.05 \times(0.1-0.05)}{0.05}\)
\(\Rightarrow e=0.5 \mathrm{~V}\)
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