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A right angled triangular loop as shown ...

A right angled triangular loop as shown below enters uniform magnetic field (at right angle to the boundary of the field) directed into the paper. Draw the graph between induced emf e and the distance along the perpendicular to the boundary of the field, (say x) along which loop moves.

A

B

C

D

Text Solution

Verified by Experts

The correct Answer is:
C
For, `0 le x le, phi=1/2 Bx^(2)`
`e=-(dphi)/(dt)=-Bav`
At `x-0, e=0`
at `x=a//2, e=-(Bav)/(2)`
at `x=x rarr a, e~~-Bav`
As `x=a,e=0`
`(phi=constant)`
For `alexle 2a`
`phi=1/2 Ba^(2)=constant`
For `2s le x le 3a, phi=B[(a^2)/(2)-(x_(1)^(2))/(2)]`
`e=((-)dphi)/(dt)= -(d)/(dt)[(B)/(2)(a^(2)-x_(1)^(2))]=Bx_(1)v`
At `x_(1)=0, e=0`
At `x_(1)=a//2, e=(B)/(2)av`
At `x_(1)=a, e=0` loop is outside the field.
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