The current required to be passed through a solenoid of 15 cm length and 60 turns in order of demagnetise a bar magnet of magnetic intensity 2.4×10³ Am⁻¹ is ___________ A.

Solution

Physics sir · 3 steps

Demagnetising means the solenoid's field intensity H = nI must equal the magnet's 2.4 × 10³.

1. What is needed. To demagnetise the bar magnet, the solenoid must produce a magnetising field equal to the given intensity, H = 2.4 × 10³ A m⁻¹. Inside a long solenoid this field is H = nI, where n is the number of turns per metre.
2. Turns per metre. The length is 15 cm = 0.15 m, so n = 60/0.15 = 400 m⁻¹.

   I = H/n = (2.4 × 10³ A m⁻¹)/(400 m⁻¹) = 6 A

3. So. The current needed is 6 A.

The common slip: 600. Taking the length in centimetres gives n = 4 and a current of 600 A. Convert 15 cm to 0.15 m first.

The idea to keep. Inside a solenoid, H = nI with n in turns per metre. Divide the total turns by the length in metres before you use it.

Answer from the NTA final answer key. Solution by PrepWiser.