A particle of charge $1.0 \times 10^{-16} .{ }^6 \mathrm{C}$ moves through a uniform magnetic field…
A particle of charge $1.0 \times 10^{-16} .{ }^6 \mathrm{C}$ moves through a uniform magnetic field $\mathbf{B}=B_0(\hat{i}+4 \hat{j}) \mathrm{T}$. The particle velocity at some instant is $\mathbf{v}=(2 \hat{i}+4 \hat{j}) \mathrm{ms}^{-1}$ and the magnetic force acting on it is $3 \times 10^{-16} \hat{k} \mathrm{~N}$. The magnitude of $B_0$ is
$1.0 \mathrm{~T}$
$2.5 \mathrm{~T}$
$0.5 \mathrm{~T}$
$0.75 \mathrm{~T}$
Solution
Force on a charged particle moving in region of a magnetic field is given by
$F=q(\mathbf{v} \times \mathbf{B})$ ...(i)
Here, $\quad q=1.0 \times 10^{-16} \mathrm{C}$
$\begin{aligned} & \mathbf{v}=(2 \hat{i}+4 \hat{j}) \mathrm{ms}^{-1}, \mathbf{B}=B_0(\hat{i}+4 \hat{j}) \mathrm{T} \\ & \mathbf{F}=3 \times 10^{-16} \hat{k} \mathrm{~N}\end{aligned}$
substituting in eq. (i), we get
$3 \times 10^{-16} \hat{k}=1 \times 10^{-16}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 0 \\ B_0 & 4 B_0 & 0\end{array}\right|$
$\begin{aligned} & \Rightarrow \quad 3 \hat{k}=\hat{i}(0)-\hat{j}(0)+\hat{k}\left(8 B_0-4 B_0\right) \\ & \Rightarrow \quad 3 \hat{k}=4 B_0 \hat{k} \text { or } B_0=\frac{3}{4}=0.75 \mathrm{~T}\end{aligned}$