Given: Number of turns, \( N = 200 \)
Area of the coil, \( A = 0.01 \, \text{m}^2 \)
Change in magnetic field, \( \Delta B = 0.5 \, \text{T} \)
Time taken for the change, \( \Delta t = 0.1 \, \text{s} \)
Step 1: Formula for Induced emf The induced emf \( \varepsilon \) in a coil due to a change in magnetic flux is given by Faraday’s Law: \[ \varepsilon = -N \frac{\Delta \Phi}{\Delta t} \] where \( \Delta \Phi = B A \) is the change in magnetic flux.
Step 2: Calculate the Induced emf The change in magnetic flux is: \[ \Delta \Phi = \Delta B \times A = 0.5 \, \text{T} \times 0.01 \, \text{m}^2 = 0.005 \, \text{T m}^2 \] Substitute into the formula for emf: \[ \varepsilon = -200 \times \frac{0.005}{0.1} = -200 \times 0.05 = -10 \, \text{V} \] The induced emf is \( 1 \, \text{V} \) (ignoring the negative sign since we are only interested in the magnitude).
Answer: The correct answer is option (a): 1 V.
Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of $\frac{1}{\sqrt{2}}$ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of $10 \mathrm{~cm} / \mathrm{s}$, induced emf between points A and E is _______ mV.}