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Question

The coercive force for a certain permanent magnet is 4.0 × 104 A m−1. This magnet is placed inside a long solenoid of 40 turns/cm and a current is passed in the solenoid to demagnetise it completely. Find the current.


Solution

Given:
Number of turns per unit length, n = 40 turns/cm = 4000 turns/m
Magnetising field, H = 4 × 104 A/m
Magnetic field inside a solenoid (B) is given by,
B = µ0nI,
where, n = number of turns per unit length.
           I = current through the solenoid.             
 Bμ0 = nI = H
H=NlII=HlN = HnI=4×1044000=10 A

Physics
HC Verma - II
Standard XII

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