Question

A charged particle having charge q experiences a force $\overrightarrow{F}=q(-\overrightarrow{j}+\overrightarrow{k})N$ in a magnetic field B when it has a velocity $v_1=1\hat{i}m/s$. The force becomes $\overrightarrow{F}=q(\overrightarrow{i}-\overrightarrow{k})N$ when the velocity is changed to $v_2=1\hat{j}m/sec$. The magnetic induction vector at that point is :

• A. $(\hat{i}+\hat{j}+\hat{k})T$
• B. $(\hat{i}-\hat{j}-\hat{k})T$
• C. $(-\hat{i}-\hat{j}+\hat{k})T$
• D. $(\hat{i}+\hat{j}-\hat{k})T$

Answer 1

The correct option is A $(\hat{i}+\hat{j}+\hat{k})T$

Let the magnetic induction at required point be

$\overline{B}=(B_x\hat{i}+B_y\hat{j}+B_z\hat{k})T$

Magnetic Force $\left(\overline{F}\right)$ on charged particle (q) moving with Velocity (V) in magnetic field $\left(\overline{B}\right)$ is given as

$\overline{F}=q(\overline{V}\times \overline{B})$

Case 1: $q(-\hat{j}+\hat{k})=q(\hat{i}\times (B_x\hat{i}+B_y\hat{j}+B_z\hat{k}))$

$-\hat{j}+\hat{k}=B_y\hat{k}-B_z\hat{j}$

$B_z=\mathrm{1....}\left(i\right)$

$B_y=+\mathrm{1....}\left(ii\right)$

Case 2: $q(\hat{j}-\hat{k})=q\left(\hat{j}(B_x\hat{i}+B_y\hat{j}+B_z\hat{k})\right)$

$\therefore \hat{j}-\hat{k}=-B_x\hat{k}+B_z\hat{i}$

$\therefore B_x=\mathrm{1....}\left(iii\right)$

From $\left(i\right),\left(ii\right)\&\left(iii\right)$

$\overline{B}=(\hat{i}+\hat{j}+\hat{k})T$ Thus, option A