Ncert Exercises & Solutions

A magnetic field $B = B_{0} \sin (ωt) \hat{k}$ covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d (figure). The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity v, what is the current in the circuit? What is the force needed to keep the wire moving at constant velocity?

Hint: The change in flux results in the current in the circuit and a force on the wire.

Step 1: Let us assume that the parallel wires are at y=0 i.e., along x-axis and y=d. At t=0, AB is at x=0, i.e., along the y-axis and moves with a velocity v. Let at time t, wire is at x =vt.

Now, the motional emf across AB is:

= (B_{0} sinωt) vd (- \hat{j})

EMF due to change in field (along OBAC)

= - B_{0} ω cosωt \times x (t) d

Step 2: Total emf in the circuit = emf due to change in field (along OBAC) + the motional emf across AB

= - B_{0} d [ωxcosωt + vsin (ωt)]

The electric current in a clockwise direction is given by; i

= \frac{B_{0} d}{R} (ωxcosωt + vsinωt)

Step 3: The force acting on the conductor is given by; F=ilBsin

90 °

=ilB

Substituting the values, we have;

Force needed along i = \frac{B_{0} d}{R} (ωxcosωt + vsinωt) \times d \times B_{0} \sin ωt

= \frac{B_{0}^{2} d^{2}}{R} (ωxcos ωt + vsinωt) sinωt

This is the required expression for force.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions