1 μC charge moving with velocity bar nu = (hat i - 2hat j + 3hat k ) m / s in the region of magnetic field bar B = (2hat i + 3hat j - 5hat k ) T . The magnitude of force acting on it is √(α)× 10 - 6 N . The value of α is

Physics · Moving Charges and Magnetic Field

JEE Main 2026 — 2 April, Morning Shift — Question 17

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

Answer: 171

Numerical answer — enter this value.

Step-by-step solution

$\bar{F} = q(\bar{v}\times\bar{B}) = 10^{-6}[( -2* -5 - 3*3)\hat{i} + ...] = 10^{-6}(\hat{i}+11\hat{j}+7\hat{k})$ → magnitude $\sqrt{1+121+49}=\sqrt{171}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Morning Shift
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Motion of Charged Particles in magnetic Fields