JEE Physics 2025 Question with Solution

Given: length of wire $L = 4.0 \, \text{cm} = 4.0 \times 10^{-2} \, \text{m}$, current $I = 8 \, \text{A}$, magnetic field $B = 0.15 \, \text{T}$, and the wire is perpendicular to the field so $\theta = 90^\circ$.

Find: the magnetic force on the wire in $\text{mN}$.

The magnetic force on a straight current-carrying wire is

$$F_m = I L B \sin\theta$$

Substituting the given values,

$$F_m = (8 \, \text{A})(4.0 \times 10^{-2} \, \text{m})(0.15 \, \text{T})\sin(90^\circ)$$

Since $\sin 90^\circ = 1$,

$$F_m = 8 \times 4.0 \times 10^{-2} \times 0.15$$ $$F_m = 8 \times 0.04 \times 0.15$$ $$F_m = 0.32 \times 0.15$$ $$F_m = 0.048 \, \text{N}$$

Now convert Newton to milliNewton using $1 \, \text{N} = 1000 \, \text{mN}$:

$$F_m = 0.048 \, \text{N} \times 1000 = 48 \, \text{mN}$$

Therefore, the magnetic force on the wire is $48 \, \text{mN}$.

Given: $I = 8 \, \text{A}$, $L = \frac{4}{100} \, \text{m}$, $B = 0.15 \, \text{T}$, and $\theta = 90^\circ$.

Find: magnetic force in $\text{mN}$.

Because the wire is perpendicular to the magnetic field, use

$$F = IlB$$

Substitute directly:

$$F = 8 \times \frac{4}{100} \times 0.15$$ $$F = 48 \times 10^{-3} \, \text{N}$$ $$F = 48 \, \text{mN}$$

Therefore, the required numerical value is 48.