1 μC charge moving with velocity v = (i - 2j + 3k) m/s in the region of magnetic field B = (2i + 3j - 5k) T. The magnitude of force acting on it is √(α) × 10⁻⁶ N. The value of α is _______. Numerical answer type. Work it out, then open the answer.

Answer: 171. v× B = ( i - 2 j + 3 k)×(2 i + 3 j - 5 k) = ((-2)(-5) - 3·3) i + (3·2 - 1(-5)) j + (1·3 - (-2)·2) k = i + 11 j + 7 k | v× B| = √(1 + 121 + 49) = √(171) F = 10⁻⁶√(171) N, so α = 171.

1 μC charge moving with velocity v = (i - 2j + 3k) m/s in the region… | JEE Main 2026 (2 Apr, Shift 1)

Q8JEE Main 2026 · 2 APR · SHIFT 1Medium

Moving Charge in a Magnetic Field

1 μC charge moving with velocity $\vec{v} = (\hat{i} - 2\hat{j} + 3\hat{k})$ m/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 $\sqrt{\alpha} \times 10^{-6}$ N. The value of $\alpha$ is _______.

Numerical answer type. Work it out, then open the answer.

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Answer: 171

$\vec v\times\vec B = (\hat i - 2\hat j + 3\hat k)\times(2\hat i + 3\hat j - 5\hat k)$
$= ((-2)(-5) - 3\cdot3)\hat i + (3\cdot2 - 1(-5))\hat j + (1\cdot3 - (-2)\cdot2)\hat k = \hat i + 11\hat j + 7\hat k$
$|\vec v\times\vec B| = \sqrt{1 + 121 + 49} = \sqrt{171}$
$F = 10^{-6}\sqrt{171}\text{ N}$ , so $\alpha = 171$ .

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