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

Let the magnetic induction at required point be 

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

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

$\bar{F}=q\left(\bar{V}\times \bar{B}\right)$

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

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

$B_z=1....\left(i\right)$

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

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

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

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

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

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