TS EAMCET 03 May 2019 Shift 2 Solved Paper

The following shows a semicircle of length L, moving with velocity v in a uniform magnetic field.
The expression for the induced emf in a conductor of length l is given by,

The expression for the induced emf in a conductor of length l is
given by
e‌‌=B(v×I)
‌‌=Bvl‌sin‌θ‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I)
The effective length of arc for induced emf is given by,
l‌eff ‌=2R
=‌

2L

π

‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II)
From equations (1) and (2)
e=Bv‌

2L

π

‌sin‌45∘(∵θ=45∘)
e=BvL‌

√2

π

‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(III)
The induced emf is given by,
φ=αBvL‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(IV)
Therefore, from equations (3) and (4)
α=‌

√2

π