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 $P Q$ ?

• A

  $\frac{B l V}{3 R}$
• B

  $\frac{B l v}{2 R}$
• C

  $\frac{3 B l v}{2 R}$
• Correct answer
  D

  $\frac{2 B l v}{3 R}$

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, $R^{\prime}=\frac{R \times R}{R+R}=\frac{R}{2}$ 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 $\varepsilon=B l v$ ...(i) Now, from figure (b), the equivalent resistance is $R_{\mathrm{eq}}=R^{\prime}+R=\frac{R}{2}+R=\frac{3 R}{2}$ Now, induced current in the circuit ( $I$ through $P Q$ ) is $I=\frac{\varepsilon}{R_{\mathrm{eq}}}=\frac{B l v}{(3 R / 2)}=\frac{2 B l v}{3 R} \quad$ [From Eq. (i)]

Hence, the correct answer is option: 'D'.