# 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. - Physics

Numerical

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?

#### Solution

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 = lb

= 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

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 × 10−4 V which lasts for 2 s.

(b) Emf developed, e = Bbv

= 0.3 × 0.02 × 0.01 = 0.6 × 10−4 V

Time taken to travel along the lenght, t = "Distance travelled"/"Velocity" = "l"/"v"

= 0.08/0.01

= 8 s

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

Concept: Inductance - Mutual Inductance
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#### APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 6 Electromagnetic Induction
Exercise | Q 4 | Page 230
NCERT Physics Part 1 and 2 Class 12
Chapter 6 Electromagnetic Induction
Exercise | Q 6.4 | Page 230

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