A rectangular wire loop of sides 8 cm an | Class 12 Physics Chapter Electromagnetic Induction, Electromagnetic Induction NCERT Solutions

Question: 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?
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 = 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 traveled / 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 length, t = Distance traveled / Velocity = l / v
                                                                     =  0.08 / 0.01 = 8 s

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


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Comments

  • Pritam Shaw
  • 2019-09-26 01:39:01

❣️


  • Saba
  • 2019-09-04 09:14:50

Where I will get remaining questions of chapter electromagnetic induction.. please tell.


  • Debanjali Dey
  • 2015-03-07 01:14:41

Really helpful


Comment(s) on this Question

Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 1: A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform....