Question

A charge q is moving with a velocity ${\overline{v}}_1$ = $1\hat{i}m/s$ at a point in a magnetic field and experiences a force ${\overline{F}}_1=q[-1\hat{j}+1\hat{k}]N$. If the charge is moving with a velocity ${\overline{v}}_2=1\hat{j}m/s$ at the same point, it experiences a force ${\overline{F}}_2=q[1\hat{i}-1\hat{k}]N$. The magnetic induction $\overline{B}$ at that point is

• A. $(\hat{i}+\hat{j}+\hat{k})Wb/m^2$
• B. $(\hat{i}-\hat{j}+\hat{k})Wb/m^2$
• C. $(-\hat{i}+\hat{j}-\hat{k})Wb/m^2$
• D. $(\hat{i}+\hat{j}-\hat{k})Wb/m^2$

Answer 1

The correct option is A $(\hat{i}+\hat{j}+\hat{k})Wb/m^2$

Force experienced by change in uniform magnetic field $\overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{A})$

Let magnetic field be $\overrightarrow{B}=x\hat{i}+y\hat{j}+x\hat{k}$

now case $1$:

$\overrightarrow{F}=q[-1\hat{j}+1\hat{k}]N...\left(1\right)$

$\overrightarrow{4}=1\hat{i}m/s$

$\overrightarrow{F}=q({\overrightarrow{V}}_1\times \overrightarrow{B}=q\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 0& 0\\ x& y& z\end{array}\right|=q[0\hat{i}-\hat{j}\left(z\right)+\hat{k}\left(y\right)]$

$\overrightarrow{F}=q(-z\hat{j}+y\hat{k})$

Comparing with $\left(1\right)\Rightarrow z=1,y=1$

Case $2$ :

$\overrightarrow{F}=q(1\hat{i}-1\hat{k})N...\left(2\right)$

$\overrightarrow{V_2}=1\hat{j}m/s$

$\overrightarrow{F_2}=q(\overrightarrow{V_2}\times \overrightarrow{B})=q\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 1& 0\\ x& 1& 1\end{array}\right|=q\left[\hat{i}\right(1)-(0)\hat{j}+x\hat{k}]$

$=q(1\hat{i}-x\hat{k})$

Comparing with $\left(2\right)\Rightarrow -x=-1\Rightarrow -x=-1\Rightarrow x=1$

Thus $B=+\hat{i}+\hat{j}+\hat{k}wb/m^2$