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JEE Main 2026 · 2 Apr · Shift 1, Physics question 48

1 μC charge moving with velocity v⃗=(i^−2j^+3k^)\vec v=\left(\hat i-2\hat j+3\hat k\right) m/s in the region of magnetic field B⃗=(2i^+3j^−5k^)\vec B=\left(2\hat i+3\hat j-5\hat k\right) T. The magnitude of force acting on it is α×10−6\sqrt\alpha\times10^{-6} N. The value of α\alpha is ________.

Answer and worked solution

Official answer: 171

Cross product.
v⃗×B⃗=∣i^j^k^1−2323−5∣=i^+11j^+7k^,∣v⃗×B⃗∣=171.\vec v\times\vec B=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-2&3\\2&3&-5\end{vmatrix}=\hat i+11\hat j+7\hat k,\qquad|\vec v\times\vec B|=\sqrt{171}.

Force. F=qvBsin⁡θ=10−6171F=qvB\sin\theta=10^{-6}\sqrt{171} N, so α=171\alpha=171.

Answer: 171

Topics:Force on a moving chargeVectors
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