A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?


Given:

Length of rectangular wire = 8 cm


Breadth of rectangular wire = 2 cm



The area of the rectangular wire can be calculated as follows:


A = l × b


A = 8 cm × 2 cm = 16 cm2


In m2, area will be A = 16 × 10 - 4 m2


Magnitude of magnetic field B = 0.3T


Velocity of the loop, v =1 cms - 1 = 0.01ms - 1


(a) The emf developed across the cut can be calculated as follows:


e = Blv …(1)


where B is the magnetic field


l is the length of the loop and,


v is the velocity of the rectangular loop


substituting the values in equation (1), we get,


e = (0.3 T × 0.08m × 0.01ms - 1)


e = 2.4 × 10 - 4 V


Time taken for travelling along the width of the loop can be calculated as follows:


T =distance travelled/velocity with which distance is travelled


T = b(width of the loop)/v (velocity)


T = 0.02 m/0.01 ms - 1


T= 2 seconds


This means that the induced voltage will last for very short direction of 2 seconds.


(b) The emf developed across the cut if the velocity of the loop is 1 cms–1 in a direction normal to the shortest side of the loop can be calculated as follows:


e = Bbv…(2)


where B is the magnetic field


b is the width of the loop and,


v is the velocity of the rectangular loop


substituting the values in equation (2), we get,


e = (0.3 T × 0.02 m × 0.01 ms - 1)


e = 0.6 × 10 - 4 V


Time taken for travelling along the width of the loop can be calculated as follows:


T =distance travelled/velocity with which distance is travelled


T = l(length of the loop)/v


T = 0.08 m/0.01ms - 1


T = 8 seconds


This means that the induced voltage will last for 8 seconds.


4
1