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Karnataka Board PUCPUC Science Class 11

Figure Shows a Long U-shaped Wire of Width L Placed in a Perpendicular Magnetic Field B. a Wire of Length L Is Slid on the U-shaped Wire with a Constant Velocity V Towards Right. - Physics

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Question

Figure shows a long U-shaped wire of width l placed in a perpendicular magnetic field B. A wire of length l is slid on the U-shaped wire with a constant velocity v towards right. The resistance of all the wires is r per unit length. At t = 0, the sliding wire is close to the left edge of the U-shaped wire. Draw an equivalent circuit diagram, showing the induced emf as a battery. Calculate the current in the circuit.

Diagram
Sum
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Solution

The induced emf is given by

e = Bvl

Total resistance, R = r × Total length of the wire

Because the length of the movable wire is l and the distance travelled by the movable wire in time t is vt, the total length of the loop is 2 (l + vt).

∴ e = i × 2r (l + vt)

Bvl = 2ri (l + vt)

\[\Rightarrow i = \frac{Bvl}{2r(l + vt)}\]

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Chapter 16: Electromagnetic Induction - Exercises [Page 308]

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HC Verma Concepts of Physics Vol. 2 [English] Class 11 and 12
Chapter 16 Electromagnetic Induction
Exercises | Q 36 | Page 308
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