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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 $P Q$ ?
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Verified Answer
The correct answer is:
$\frac{2 B l v}{3 R}$
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)]
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)]
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