Question

A conducting wire bent but in the shape of semicircle has length $L$ and moves in its plane with constant velocity $v$. A uniform magnetic field $B$ exists in the direction perpendicular to the plane of the wire. The velocity makes an angles $45^{\circ}$ to diameter joining free ends and the emf induced between the ends of the wire is $\Phi=\alpha(B v L)$ The value of constant $\alpha$ is

• A. $\sqrt{2}$
• B. $\frac{2}{\pi}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{\sqrt{2}}{\pi}$

Answer 1

Answer: (D)

A semicircle of length $L$, moving with velocity vin a uniform magnetic field is shown in the figure.

As, we know that the induced emf in a conductor of length $l$,
$e=B( v \times l )$
$=B v l \sin \theta$ ...(i)
Since, we have semi-circular arc in this problem. So, the effective length of arc for induced emf,
$l_{ eff }=2 R=\frac{2 L}{\pi}$
$(\because L=\pi R)$ ...(ii)
From Eqs. (i) and (ii), we get
$e=B v \frac{2 L}{\pi} \sin 45^{\circ}$
$\left(\because \theta-45^{\circ}\right)$
$\Rightarrow e=B v L \frac{\sqrt{2}}{\pi}$ ...(iii)
Now, given induced emf, $\phi=\alpha B v L$ ...(iv)
Hence, from Eqs. (iii) and (iv), we get
$\alpha=\frac{\sqrt{2}}{\pi}$