JEE Main 2023 — Magnetic Properties Of Matter Question with Solution

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 ($2.4\times 10^3{\text{Am}}^{-1}$), you need to create a magnetizing field in the solenoid that is equal in magnitude but opposite in direction.

Given:

• ($H=2.4\times 10^3{\text{Am}}^{-1}$)
• (N = 60) turns
• ($L=15\text{cm}=0.15\text{m}$) (since $(1\text{m}=100\text{cm})$)

You can rearrange the formula for (H) to solve for (I) :

$I=\frac{H\cdot L}{N}$

Substituting the given values, you get :

$I=\frac{(2.4\times 10^3{\text{Am}}^{-1})\cdot 0.15\text{m}}{60}$

$I=6\text{Amperes}$

So, the current required to be passed through the solenoid to demagnetize the bar magnet is (6 $\text{A}$).