Question

A conductor (shown in the figure) carrying constant current I is kept in the x-y plane in a uniform magnetic field $\overrightarrow{B}$. If F is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s) is (are)

• A. if $\overrightarrow{B}$ is along $\hat{z},F\propto (L+R)$
• B. if $\overrightarrow{B}$ is along $\hat{x},F=0$
• C. if $\overrightarrow{B}$ is along $\hat{y},F\propto (L+R)$
• D. if $\overrightarrow{B}$ is along $\hat{z},F=0$

Answer 1

Answer: (A), (B), (C)

The correct options are
A if $\overrightarrow{B}$ is along $\hat{z},F\propto (L+R)$
B if $\overrightarrow{B}$ is along $\hat{x},F=0$
C if $\overrightarrow{B}$ is along $\hat{y},F\propto (L+R)$

The vector sum of all current element in wires not parallel to x axis will lead to resultant current element of length R and current along x axis. Total length of current element:
$dl=2(L+R)\hat{x}$
$\overrightarrow{F}=i\left(2\right(L+R\left)\hat{x}\right)\times \overrightarrow{B}$
If $\overrightarrow{B}$ along $\hat{z}$ then $F$ will be along $-\hat{y}$ and will be propotional to (L+R)
If $\overrightarrow{B}$ along $\hat{x}$ then $F=0$ since vector product is zero.
If $\overrightarrow{B}$ along $\hat{y}$ then $F$ will be along $\hat{z}$ and will be prop to (L+R)
If $\overrightarrow{B}$ along $\hat{z}$ then $F$ will be along $-\hat{y}$ and will not be equal to 0.