Shows a square loop having 100 turns, an area of `2.5 X10^(-3) m^2` and a resistance of `100 Omega`. The magnetic field has a magnitude B =0.40 T. Find the work done in pulling the loop out of the field, slowly and uniformly is 1.0 s.

Explanation

The side of the square is ` l = sqrt(2
5 X10^(-3) m^2) = 0
05m
As it is uniformly pulled out in 1
0 s, the speed of the loop is `v = 0
05 ms^(-1)`
The emf induced in the left arm of the loop is `epsilon = NvBl` = `100 X (0
05 ms^(-1))`
The emf induced in the left arm of the loop is `epsilon = NvBl` `= 100 X(0
05 ms^(-1) X (0
40 T) X (0
05m)` =0
1V
The current in the loop is ` i =(0
1 V)/(100 Omega) =1
0 X 10 ^(-3) A`
The force on the left arm due to the magnetic field is ` F = NilB = 100X (1
0X10^(-3)A) X (0
05m)X (0
40 T)` =2
0 X 10^(-3) N
The force is towards left in the figure
To pull the loop uniformly, an external force of `2
0 X 10^(-3) N towards right must be applied
the work done by this force is` W= (2
0 X 10^(-3)N)X (0
05 m) = 1
0 X 10 ^(-4) J