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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.
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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.
\[\therefore \frac{B}{\mu_0} = nI = H\]
\[\Rightarrow H = \frac{N}{l}I\]
\[ \Rightarrow I = \frac{Hl}{N} = \frac{H}{n}\]
\[ \Rightarrow I = \frac{4 \times {10}^4}{4000} = 10 A\]
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