The magnetic field existing in a region is given by `vecB = B_0(1 + x/1)veck` . 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.

#### Solution

Given:

Magnetic field, `vecB = B_0(1 + x/1)veck`

Length of the edge of a square loop = *l*

Electric current flowing through it = *i*

As per the question, the loop is placed with its edges parallel to the *X*−*Y* axes.

In the figure, arrow denotes the direction of force on different sides of the square.

The net magnetic force experienced by the loop,

`vecF = ivecl xx vecB`

Force on AB:

Consider a small element of length dx at a distance x from the origin on line AB.

Force on this small element,

dF = iB_0 on the full length of AB,

F_{AB }= \[\int\limits_{x=0}^{x=0}\] iB_0 `(1 + x/l)`

= `iB_0` \[\int\limits_{x=0}^{x=0}\] `(dx + 1/l xdx)`

= `iB_0(l + 1/2)`

= `(3iBgl)/(2)`

Force on AB will be acting downwards.

Similarly, force on CD,

`F_2 = iB_0 (l + l/2)`

`=(3iBgl)/(2)`

Force on AB will be acting downwards.

Similarly, force on CD,

`F_2 = iB_0 (l + 1/2)`

= `(3iBgl)/2`

So, the net vertical force =* **F*_{1}_{ }−* **F*_{2} = 0

Force on AD,

`F_4 = iB_0l (1 + 1/l)`

= 2iB_{0}l

Force on BC

`F_4 = iB_0l(1 + 1/l)`

=2iB_{0}l

So, the net horizontal force =* **F*_{4}−*F*_{3} = *iB*_{0}*l*