JEE Main 2026 Apr 2 Shift 1, Physics Q23: Motion of a Charged Particle in a Magnetic Field

The Lorentz force is $\vec{F}=q(\vec{v}\times \vec{B})$. Computing the cross product:

$\vec{v}\times \vec{B}=\left|\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ 1-23\\ 23-5\end{array}\right|=\hat{i}((-2)(-5)-(3)(3))-\hat{j}((1)(-5)-(3)(2))+\hat{k}((1)(3)-(-2)(2))$

$=\hat{i}(10-9)-\hat{j}(-5-6)+\hat{k}(3+4)=\hat{i}+11\hat{j}+7\hat{k}$

Magnitude $=\sqrt{1^2+11^2+7^2}=\sqrt{1+121+49}=\sqrt{171}$

With $q=1\mu$C $=10^{-6}$ C: $|\vec{F}|=\sqrt{171}\times 10^{-6}$ N, so $\alpha =171$.