Question 139: Magnetism and Matter

Question

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 $$2.4	imes10^3~Am^{-1}$$ is ___________ A.

Accepted Answer

We know that the magnetizing field (H) inside a solenoid is given by the formula :
$ H = \frac{N \cdot I}{L} $
where (N) is the number of turns, (I) is the current in Amperes, and (L) is the length of the solenoid in meters.
To demagnetize a bar magnet that has a magnetic intensity (H) of ($2.4 	imes 10^3 \, 	ext{Am}^{-1}$), you need to create a magnetizing field in the solenoid that is equal in magnitude but opposite in direction.
Given:
($H = 2.4 	imes 10^3 \, 	ext{Am}^{-1}$)
(N = 60) turns
($L = 15 \, 	ext{cm} = 0.15 \, 	ext{m}$) (since $(1\, 	ext{m} = 100 \, 	ext{cm})$)
You can rearrange the formula for (H) to solve for (I) :
$ I = \frac{H \cdot L}{N} $
Substituting the given values, you get :
$ I = \frac{(2.4 	imes 10^3 \, 	ext{Am}^{-1}) \cdot 0.15 \, 	ext{m}}{60} $
$ I = 6 \, 	ext{Amperes} $
So, the current required to be passed through the solenoid to demagnetize the bar magnet is (6 $\, 	ext{A}$).

Magnetism and Matter

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 $2.4\times10^3~Am^{-1}$ is ___________ A.

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