Question

In the given diagram, force acting on the conductor $\text{OA}$, aligned along the body digonal of the cube, is:
( Given, side of the cube is $2\text{m},I=2\text{A},\overrightarrow{B}=3\hat{k}\text{T}$ )

• A. $6(\hat{i}+\hat{j})\text{N}$
• B. $3(\hat{i}+\hat{j})\text{N}$
• C. $(\hat{i}+\hat{j})\text{N}$
• D. $12(\hat{i}+\hat{j})\text{N}$

Answer 1

The correct option is D $12(\hat{i}+\hat{j})\text{N}$

As the side of the cube is $2\text{m}$
$\therefore$ for the given conductor,

$\overrightarrow{l}=2\hat{i}+2\hat{j}+2\hat{k}$

Now, ${\overrightarrow{F}}_m=I(\overrightarrow{l}\times \overrightarrow{B})$

$=2\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 2& 2\\ 0& 0& 3\end{array}\right]$

$=2[6\hat{i}+6\hat{j}]$

$\therefore {\overrightarrow{F}}_m=12(\hat{i}+\hat{j})\text{N}$

Hence, option $\left(D\right)$ is the correct answer.