1.

A conducting square frame of side 'a' and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity 'V'. The emf induced in the frame will be proportional to :

Answer»

`1/x^2`
`1/(2x-a)^2`
`1/"(2x+a)"`
`(1/((2x-a))-1/(2x+a))`

Solution :INSIDE (1) magnitude of induced emf is `epsilon_1=B_1 Vl`
Inside (2) magnitude of induced emf is `epsilon_2=B_2 Vl`
Induced emf value in frame `EPSILON=epsilon_1-epsilon_2` [Side (1) is NEAR to wire and `B PROP 1/r`]
`THEREFORE epsilon=Vl (B_1-B_2)`
`therefore epsilon prop (B_1-B_2)`
but `B_1 prop 1/(x-a/2)` and `B_2 prop 1/(x+a/2)`
`therefore epsilon prop (1/(x-a/2)-1/(x+a/2))`
`therefore epsilon prop (2/(2x-a) -2/(2x+a))`
`therefore epsilon prop (1/(2x-a)-1/(2x+a))` ( `because` 2 constant)


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