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