Question

A charged particle q is moving with a velocity $\overrightarrow{v_1}=2\hat{i}m/s$ at a point in a magnetic field $\overrightarrow{B}$ and experiences a force $\overrightarrow{F_1}=q(\hat{k}-2\hat{j})N.$ If the same charge moves with velocity $\overrightarrow{v_2}=2\hat{j}m/s$ from the same point in that magnetic field and experiences a force $\overrightarrow{F_2}=q(2\hat{i}+\hat{k})N$, the magnetic induction at that point will be :

• A. $\hat{i}+\frac{1}{2}\hat{j}-\frac{1}{2}\hat{k}$
• B. $-\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}+\hat{k}$
• C. $\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}+\hat{k}$
• D. $-\frac{1}{2}\hat{i}+\hat{j}+\frac{1}{2}\hat{k}$

Answer 1

Answer: (B)

$-\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}+\hat{k}$

Let the magnetic induction at required point be $\overline{B}=B_x\hat{i}+B_y\hat{j}+B_z\hat{k}$

Magnetic force $\left(\overline{F}\right)$ on particle of charge (q), moving with velocity $\overline{V}$ in the magnetic field region of intensity $\left(\overline{B}\right)$ is given as

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

Case 1: $\overline{F_1}=q(\overline{V_1}\times \overline{B})$

$q(\hat{k}-2\hat{j})=q(2\hat{i}\times (B_x\hat{i}+B_y\hat{j}+B_z\hat{k}))$

$\hat{k}-2\hat{j}=2B_y\hat{k}-2B_z\hat{j}$

$\therefore 2B_y=1,B_y=1/2.....\left(i\right)$

$-2=-2B_z,\therefore B_z=\mathrm{1.....}\left(ii\right)$

Case 2: $\overline{F_2}=q(\overline{V_2}\times \overline{B})$

$q(2\hat{i}+\hat{k})=q(2\hat{j}\times (B_x\hat{i}+B_y\hat{j}+B_z\hat{k}))$

$2\hat{i}+\hat{k}=-2B_x\hat{k}+2B_z\hat{i}$

$\therefore -2B_x=1,\therefore +B_x=1/2......\left(iii\right)$

Thus from $\left(i\right),\left(ii\right)\&\left(iii\right)$

$\overline{B}=-\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}+\hat{k}$