JEE Physics 2023 Question with Solution

Given: Magnetic intensity of the bar magnet is $H = 2.4 \times 10^3 \, \text{A/m}$, number of turns $N = 60$, and length of the solenoid $l = 15 \, \text{cm} = 0.15 \, \text{m}$.

Find: The current $I$ required to demagnetize the bar magnet.

For a solenoid, the magnetic intensity is

$$H = \frac{NI}{l}$$

Rearranging for $I$,

$$I = \frac{Hl}{N}$$

Substitute the values:

$$I = \frac{(2.4 \times 10^3)(0.15)}{60}$$

Simplifying,

$$I = \frac{360}{60} = 6 \, \text{A}$$

Therefore, the current required is $6 \, \text{A}$.

Given: $H = 2.4 \times 10^3 \, \text{A/m}$, $N = 60$, $l = 0.15 \, \text{m}$.

Find: Current $I$.

1. Use the relation for magnetic intensity inside a solenoid:

$$H = \frac{NI}{l}$$

2. Make $I$ the subject:

$$I = \frac{Hl}{N}$$

3. Substitute the given values:

$$I = \frac{(2.4 \times 10^3)(0.15)}{60}$$

4. Multiply the numerator:

$$(2.4 \times 10^3)(0.15) = 360$$

5. Divide by $60$:

$$I = \frac{360}{60} = 6$$

Hence, the required current is $6 \, \text{A}$.