KCET · Physics · Electromagnetic Induction
A uniform magnetic field of strength \(B=2 \mathrm{mT}\) exists vertically downwards. These magnetic field lines pass through a closed surface as shown in the figure. The closed surface consists of a hemisphere \(S_1\), a right circular cone \(S_2\) and a circular surface \(S_3\). The magnetic flux through \(S_1\) and \(S_2\) are respectively.
- A \(\Phi_{S_1}=-20 \mu \mathrm{~Wb}, \Phi_{S_2}=+20 \mu \mathrm{~Wb}\)
- B \(\Phi_{S_1}=+20 \mu \mathrm{~Wb}, \Phi_{S_2}=-20 \mu \mathrm{~Wb}\)
- C \(\Phi_{S_1}=-40 \mu \mathrm{~Wb}, \Phi_{S_3}=+40 \mu \mathrm{~Wb}\)
- D \(\Phi_{S_1}=+40 \mu \mathrm{~Wb}, \Phi_{S_2}=-40 \mu \mathrm{~Wb}\)
Answer & Solution
Correct Answer
(A) \(\Phi_{S_1}=-20 \mu \mathrm{~Wb}, \Phi_{S_2}=+20 \mu \mathrm{~Wb}\)
Step-by-step Solution
Detailed explanation
Given, \(B=2 \mathrm{mT}\)
\(=2 \times 10^{-3} \mathrm{~T}, R=\frac{10}{\sqrt{\pi}} \mathrm{~cm}\)
Magnetic flux passing through surface \(S_1\),
\(\phi_{S_1}=B S_1 \cos 180^{\circ}\)
\(=2 \times 10^{-3} \times \pi R^2(-1)\)
\(=-2 \times 10^{-3} \times \pi \times \frac{100}{\pi} \times 10^{-4}\)
\(=-20 \times 10^{-6} \mathrm{~Wb}=-20 \mu \mathrm{~Wb}\)
Since, Total entering magnetic flux \(=\) Total leaving magnetic flux
\(\therefore \phi_{S_2}=-\phi_{S_1}=20 \mu \mathrm{~Wb}\)
\(=2 \times 10^{-3} \mathrm{~T}, R=\frac{10}{\sqrt{\pi}} \mathrm{~cm}\)
Magnetic flux passing through surface \(S_1\),
\(\phi_{S_1}=B S_1 \cos 180^{\circ}\)
\(=2 \times 10^{-3} \times \pi R^2(-1)\)
\(=-2 \times 10^{-3} \times \pi \times \frac{100}{\pi} \times 10^{-4}\)
\(=-20 \times 10^{-6} \mathrm{~Wb}=-20 \mu \mathrm{~Wb}\)
Since, Total entering magnetic flux \(=\) Total leaving magnetic flux
\(\therefore \phi_{S_2}=-\phi_{S_1}=20 \mu \mathrm{~Wb}\)
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