Concepts Of Physics MCQ Edition [Volume 2]PhysicsElectromagnetic Induction
A circular coil of N turns and radius a is connected to a battery of emf E through a rheostat. The rheostat has a total length L and resistance R . The resistance of the coil is r . A small circular loop of radius a' and resistance r' is positioned coaxially with the coil. The centre of the loop is located at a distance x from the centre of the coil. Initially, the sliding contact of the rheostat is at the left end,
Options
- A₀ N a^2 a'^2 E R v 2 L (a^2 + x^2)^ 3/2 (R/2 + r)^2
- B₀ N a^2 a'^2 E R v L (a^2 + x^2)^ 3/2 (R/2 + r)^2
- C₀ N a^2 a'^2 E R v 2 L (a^2 + x^2)^ 3/2 (R + r)^2
- D₀ N a^2 a'^2 E R v 2 L (a^2 + x^2)^ 3/2 (R/2 + r)
Correct answer
A. ₀ N a^2 a'^2 E R v 2 L (a^2 + x^2)^ 3/2 (R/2 + r)^2
Step-by-step solution
Let y be the distance moved by the sliding contact from the left end of the rheostat. Since it moves at a constant speed v , we have dy dt = v . From the circuit diagram, the right end of the rheostat is connected to the battery. Therefore, the length of the rheostat included in the circuit is L - y , and its resistance is R_ rheo = R L (L - y) . The total equivalent resistance of the circuit is: R_ eq = r + R_ rheo = r + R - R L y The current in the coil is: I = E R_ eq = E r + R - R L y The rate of change of curr