A conducting wire bent but in the shape of semicircle has length L and mvoes 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^@` 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

Options

A. `sqrt2`
B. `2/ pi`
C. `1/sqrt2`
D. `sqrt2/pi`

Answer

D

Explanation

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

As , we know that the induced emf in a conductor of length l ,
`e = B(Vxx 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) = 2R = (2L)/pi (because L = pi R) ` ......(ii)
From Eqs. (i) and (ii) we get
`e = Bv (2L)/pi sin 45^@ ( because theta = 45^@)`
`rArr e Bv L sqrt2/pi ` .......(iii)
Now, given induced emf `phi = alpha B v L` .......(iv)
Hence, from Eqs. (iii) and (iv) we get
`alpha = sqrt2/pi`