Question

A charge $q$ is moving with a velocity $\overrightarrow{v_1}=\left(1\hat{i}\right)\text{m}/\text{s}$ at a point in a magnetic field and experiences a force $\overrightarrow{F_1}=q(-\hat{j}+\hat{k})$$\text{N}$. If the same charge is moving with a velocity $\overrightarrow{v_2}=\left(1\hat{j}\right)\text{m}/\text{s}$ at the same point, it experiences a force $\overrightarrow{F_2}=q(\hat{i}-\hat{k})\text{N}.$ The magnetic field $\overrightarrow{B}$ at that point is -

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

Answer 1

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

Given:

$v_1=\left(1\hat{i}\right)\text{m}/\text{s}{v}_2=\left(1\hat{j}\right)\text{m}/\text{s}{F}_1=q(-\hat{j}+\hat{k})\text{N}{F}_2=q(\hat{i}-\hat{k})\text{N}$

Let the magnetic field at the point be,

$\overrightarrow{B}=(x\hat{i}+y\hat{j}+z\hat{k})\text{Wb}/{\text{m}}^2...\left(1\right)$

Force experienced by charge particle $q$ moving with velocity$\overrightarrow{v}$ in the magnetic field $\overrightarrow{B}$ is given by $$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$$$\Rightarrow \overrightarrow{F_1}=q(\overrightarrow{v_1}\times \overrightarrow{B})$

$\Rightarrow q(-1\hat{j}+1\hat{k})=q\left[\right(1\hat{i})\times (x\hat{i}+y\hat{j}+z\hat{k}\left)\right]$

$\Rightarrow -1\hat{j}+1\hat{k}=y\hat{k}-z\hat{j}$

Comparing $LHS$ and $RHS$ we get,

$y=1$ and $z=1$

Substituting the values of $y$ and $z$ in Eq. $\left(1\right)$, we get,

$\overrightarrow{B}=x\hat{i}+\hat{j}+\hat{k}...\left(2\right)$

Similarly,

$\overrightarrow{F_2}=q(\overrightarrow{v_2}\times \overrightarrow{B})$

$\Rightarrow q(\hat{i}-\hat{k})=q\left[\right(1\hat{j})\times (x\hat{i}+y\hat{j}+z\hat{k}\left)\right]$

$\Rightarrow \hat{i}-\hat{k}=-x\hat{k}+z\hat{i}$

Comparing $LHS$ and $RHS$, we get,

$x=1$

Substituting the value of $x$ in Eq. $\left(2\right)$, we get,

$\overrightarrow{B}=(\hat{i}+\hat{j}+\hat{k})\text{Wb}/{\text{m}}^2$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(A\right)$ is the correct answer.