Q 12-04-045JEE MainJEE Main 2026 (2 Apr, Shift 1)Easy
$1\ \mu$C 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 $\sqrt\alpha\times10^{-6}$ N. The value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 171
$\vec F=q(\vec v\times\vec B)$.
$$\vec v\times\vec B=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-2&3\\2&3&-5\end{vmatrix}=\hat i(10-9)-\hat j(-5-6)+\hat k(3+4)=\hat i+11\hat j+7\hat k$$
$|\vec v\times\vec B|=\sqrt{1+121+49}=\sqrt{171}$
$|\vec F|=10^{-6}\sqrt{171}$ N, so $\alpha=171$.
Solution by Sreeraj P, M.Sc Physics