The magnetic field existing in a region is given by `vecB =B_0(1+(x)/(l))vec k.` A square loop of edge l and carrying a current I, is placed with its edges parallel to the x-y axes. Find the magnitude of the net magnetic force experienced by the loop.

Options

A. (a) `(v_(0)B_(0)d^(2))/(a)`
B. (b) `(v_(0)B_(0)d^(3))/(a^(2))`
C. ( c) `v_(0)B_(0)d`
D. (d) zero

Answer

A

Explanation

(a) Magnetic field at sid e(1): `B_(1) = B_(0)`
induced emf in (1): `e_(1) = B_(0)v_(0)d`

Magnetic field at side (2): `(B_(2) = B_(0)[1 + (d)/(a)]`
Induced emf in `3 and 4`will be zero.
Net emf: `e = e_(2) - e_(1) = (B_(0)v_(0)d^(2))/(a)`
Alternate mathod:
Let at any instant, the loop is at distance `y` as shown below.
Let us take anticlockwise direction to be positive.

flux through a strip of width `dx`: `dphi = B_(0)[1 + (x)/(a)](d)(dx)`
Net flux: `phi = B_(0)dint_(y)^(y + d)(1 + (x)/(a))`
`rarr` `phi = B_(0)d[x + (x^(2))/(2a)]_(y)^(y + d)`
`rarr` `phi = B_(0)d[d + (1)/(2a)[(y + d)^(2) - y^(2)]]`
`rarr` `phi = B_(0)d[d + (1)/(2) - d(2y + d)]`
`e = - (dphi)/(dt) = - B-(0)d[(1)/(2a) - d(2y + d)]`
`rarr` `e = - (B_(0)d^(2))/(a) (dy)/(dt) = - (B_(0)d^(2))/(a)v_(0)`
Negative sign indicates that emf induced is clockwise.