AP EAMCET · PHYSICS · Electromagnetic Induction
A straight conductor of length \(4 \mathrm{~m}\) moves at a speed of \(10 \mathrm{~m} / \mathrm{s}\). When the conductor makes on angle of \(30^{\circ}\) with the direction of magnetic field of induction of \(0.1 \mathrm{~Wb}-\mathrm{m}^2\), then induced emf is
- A \(8 \mathrm{~V}\)
- B \(4 \mathrm{~V}\)
- C \(1 \mathrm{~V}\)
- D \(2 \mathrm{~V}\)
Answer & Solution
Correct Answer
(D) \(2 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Induced emf is given by \(\begin{aligned} e & =B v l \sin \theta \\ & =0.1 \times 10 < 4 \sin 30=2 \mathrm{~V}\end{aligned}\)
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