QJEEM23Ph62a9: JEE Main 2023 Physics - Class 12 JEEM23Ph62a9 Solution: The current required to be passed through a solenoid of 15 cm…

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 $\, \text{A}$ ).