Concepts Of Physics MCQ Edition [Volume 2]PhysicsElectromagnetic Induction
As shown in the figure, 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 rheos
Options
- A₀ N a^2 a'^2 E R v 2 L (a^2 + x^2) (R + r)^2
- B₀ N a^2 a'^2 E R v L (a^2 + x^2)^ 3/2 (R + r)^2
- C₀ N a^2 a'^2 E R v 2 L (a^2 + x^2)^ 3/2 (R + r)
- D₀ N a^2 a'^2 E R v 2 L (a^2 + x^2)^ 3/2 (R + r)^2
Correct answer
D. ₀ N a^2 a'^2 E R v 2 L (a^2 + x^2)^ 3/2 (R + r)^2
Step-by-step solution
The resistance of the rheostat in the circuit at time t is given by R_ rh (t) = R L (L - vt) , since the sliding contact starts at the left end and moves right, decreasing the length of the resistive wire in the circuit. The total resistance of the circuit is R_ eq (t) = r + R - R v L t . The current in the primary coil is I(t) = E r + R - R v L t . Differentiating the current with respect to time, we get: dI dt = E ( R v L ) ( r + R - R v L t )^2 At t = 0 , the rate of change of current is: . dI dt |_ t=0 = E R v