Question

Question 6.4: 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?

Class 12 - Physics - Electromagnetic Induction Electromagnetic Induction Page-230

Answer 1

length of rectangular loop, l = 8cm = 0.08m
width of rectangular loop , b = 2cm = 0.02m
area of loop, A = l × b = 0.08 × 0.02 = 1.6 × 10^-3 m²
magnetic field strength, B = 0.3T
velocity of the loop, v = 1 cm/s = 0.01 m/s

(a) emf developed in the loop is given as
e = Blv
= 0.3 × 0.08 × 0.01 = 2.4 × 10^-4V
time taken to travel along the width,t = distance travelled/velocity = b/v
= 0.02/0.01 = 2 s

hence, the induced voltage is 2.4 × 10^-4V which lasts for 2s

(b) emf developed , e = Bbv
= 0.3 × 0.02 × 0.01 = 0.6 × 10^-4V
time taken to travel along length , t' = l/v
= 0.08/0.01 = 8s
hence,the induced voltage is 0.6 × 10^-4V which lasts for 8s

Answer 2

Answer:

Length of the rectangular wire, l = 8 cm = 0.08 m

Width of the rectangular wire, b = 2 cm = 0.02 m

Hence, area of the rectangular loop,

A = l ×b

= 0.08 × 0.02

= 16 × 10 -4 m2

Magnetic field strength, B = 0.3 T

Velocity of the loop, v = 1 cm/s = 0.01 m/s

A.Emf developed in the loop is given as:

e = Blv

= 0.3 × 0.08 × 0.01 = 2.4 × 10-4 V

Let time taken to travel along the width be t,

t=DistancetravelledVelocity=bv =0.020.01=2s

Hence, the induced voltage is 2.4 × 10 -4 V which lasts for 2 s.

B.Emf developed, e = Bbv

= 0.3 × 0.02 × 0.01 = 0.6 × 10-4 V

Let time taken to travel along the width be t,

t=DistancetravelledVelocity=lv =0.080.01=8s

Hence, the induced voltage is 0.6 × 10 -4 V which lasts for 8 s.

Explanation: