$1 \mu \mathrm{C}$ charge moving with velocity $\overrightarrow{\mathrm{v}}=(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \mathrm{m} / \mathrm{s}$ in the region of magnetic field $\vec{B}=(2 \hat{i}+3 \hat{j}-5 \hat{k})$ T. The magnitude of force acting on it is $\sqrt{\alpha} \times 10^{-6} \mathrm{~N}$. The value of $\alpha$ is ____.

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Magnetic Effects of Current

1 μ C

charge moving with velocity

v = (\hat{i} - 2 \hat{j} + 3 \hat{k}) m / s

in the region of magnetic field

B = (2 \hat{i} + 3 \hat{j} - 5 \hat{k})

T. The magnitude of force acting on it is

α \times 1 0^{- 6} N

. The value of

α

is ____.

Answer: 171

Solution

The magnetic force is

F = q (v \times B)

. The cross product is

v \times B = \hat{i} [(- 2) (- 5) - (3) (3)] - \hat{j} [(1) (- 5) - (3) (2)] + \hat{k} [(1) (3) - (- 2) (2)] = \hat{i} + 11 \hat{j} + 7 \hat{k}

. Hence

F = 1 0^{- 6} (\hat{i} + 11 \hat{j} + 7 \hat{k})

N, with magnitude

∣ F ∣ = 1 0^{- 6} 1^{2} + 1 1^{2} + 7^{2} = 171 \times 1 0^{- 6}

N. The value of

α

171

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Integer answer

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