A rectangular loop circuit has a sliding wire P Q as shown in the figure. The loop is placed in a magnetic field B , perpendicular to its plane. The resistance of the wire P Q is R . If the wire moves with constant velocity v , then find the current flowing in the wire P Q ?

Step-by-step solution

Given, P Q arm is moving with constant speed v perpendicular to the magnetic field B . Here, both resistances are in parallel, so their equivalent resistance, R^ = R R R+R = R 2 Now, the circuit becomes as shown in Fig. (b). Length of P Q=l Using expression of induced emf across P Q is given by =B l v ...(i) Now, from figure (b), the equivalent resistance is R_ eq =R^ +R= R 2 +R= 3 R 2 Now, induced current in the circuit ( I through P Q ) is I= R_ eq = B l v (3 R / 2) = 2 B l v 3 R [From Eq. (i)]