Physics · 2026
JEE Main · 2 April 2026, Shift 1 · Q48
1 μ C charge moving with velocity v =( i -2 j +3 k ) m / s in the region of magnetic field B =(2 i +3 j -5 k ) T. The magnitude of force acting on it…
$\displaystyle 1 \mu \mathrm{C}$ charge moving with velocity $\displaystyle \vec{v}=(\hat{i}-2 \hat{j}+3 \hat{k}) \mathrm{m} / \mathrm{s}$ in the region of magnetic field $\displaystyle \overrightarrow{\mathrm{B}}=(2 \hat{i}+3 \hat{j}-5 \hat{k}) \mathrm{T}$. The magnitude of force acting on it is $\displaystyle \sqrt{\alpha} \times 10^{-6} \mathrm{~N}$. The value of $\displaystyle \alpha$ is
$\displaystyle \_\_\_\_$.
Official answer
From NTA’s final answer key for this paper.
171
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.