Question

A electron experiences a force $(4.0\hat{i}+3.0\hat{j})\times {10}^{-13}N$ in a uniform magnetic field when its velocity is $2.5\hat{k}\times {10}^7{ms}^{-1}$. When the velocity becomes $(1.5\hat{i}-2.0\hat{j})\times {10}^7{\text{ms}}^{-1}$, the magnetic force of the electron is zero. The magnetic field vector $\overrightarrow{B}$ is:

• A. $-0.075\hat{i}+0.1\hat{j}$
• B. $0.1\hat{i}+0.075\hat{j}$
• C. $0.075\hat{i}-0.1\hat{j}+\hat{k}$
• D. $0.075\hat{i}-0.1\hat{j}$

Answer 1

The correct option is A $-0.075\hat{i}+0.1\hat{j}$

Let $\overrightarrow{B}=(B_x\hat{i}+B_y\hat{j}+B_z\hat{k})$
$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$
Given,
$(4\hat{i}+3\hat{j})\times {10}^{-13}=(-1.6\times {10}^{-19})[2.5\hat{k}\times (B_x\hat{i}+B_y\hat{j}+B_z\hat{k})]\times {10}^7$
$0.4\hat{i}+0.3\hat{j}=4B_y\hat{i}-4B_x\hat{j}$
$\Rightarrow B_y=0.1,B_x=-0.075$
Also, since $\overrightarrow{F}=0$ when $(1.5\hat{i}-2\hat{j})\times {10}^7{\text{ms}}^{-1}$, $\overrightarrow{B}$ must be in the $x-y$ plane and hence $B_z=0$. Thus,
$\overrightarrow{B}=-0.075\hat{i}+0.1\hat{j}$