A square loop of side 12 cm with its side parallel to x and y axis move with velocity of 8 cms^-1in 
the positive x direction in an environment containing a magnetic field in this positive z direction.The 
field is neither uniform in space nor constant in time.It has a grdient of 10^-3 Tcm^-1 along the negative x direction (i.e., it increases by 10^-3 Tcm^-1 along the negative x direction i.e.it increases 
by 10^-3 Tper cm as one moves in the -ve x direction),and it is decreasing in time at the rate 10^-3
Ts^-1.Determine the direction and magnitude of the induced current in the loop if its resistance is 4.5
m ohm.

Asked by Balbir | 26th Jul, 2017, 09:03: PM

Expert Answer:

The following steps and pointers need to be followed for solving the query:
 
1) First, find the area of the loop as it is inside the field. A = 144 × 10-4 m2.
 
2) Consider the time variation of the field. It is 10-3 Ts-1 With this, we get the change of flux due to time variation as (dΦ/dt)t = εt = (B/t)A = 10-3 Ts-1 × 144 × 10-4.
 
3) Flux due to motion of loop in non-uniform field is (dΦ/dt)v = (B/t)Av = εv.
 
4) Add both the values of emf to get the total induced emf.
 
5) Find the induced current by I = ε/R.
 
6) Note that the direction of induced current will be such as to increase the current through +z direction.
 

Answered by Romal Bhansali | 28th Jul, 2017, 01:43: PM