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An equilateral triangular loop ADC havin...

An equilateral triangular loop `ADC` having some resistance is pulled with a constant velocity `v` out of a uniform magnetic field directed in to the paper. At time `t = 0`, side `DC` of the loop at is at edge of the magnetic field.

The induced current `(i)` versus time `(t)` graph will be as

A

(a)

B

(b)

C

( c)

D

(d)

Text Solution

Verified by Experts

The correct Answer is:
B
(b) Let `2a` be the side of the triangle and `b` be the length `AE`.

`(AH)/(AE) = (GE)/(EC)` `rarr` `GH = ((AH)/(AE))EC`
or `GH = ((b - vt))/(b).a = (a - (a)/(b)vt)`
`:.` `FG = 2GH = 2[a - (a)/(b)vt]`
`:.` Induced emf, `e = Bv(FG) = 2Bv (a - (a)/(b)vt)`
`:.` Induced current, `i = (e)/(R) = (2Bv)/(R )[a - (a)/(b)vt]`
or `i = k_(1) - k_(2)t`
Thus `i - t` graph is a straight line with nagative slope and positive intercept.
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