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.

Answer 1

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.

\[\therefore   \frac{B}{\mu_0}   =   nI   =   H\]

\[\Rightarrow H = \frac{N}{l}I\]

\[ \Rightarrow I = \frac{Hl}{N}   =   \frac{H}{n}\]

\[ \Rightarrow I = \frac{4 \times {10}^4}{4000} = 10  A\]