Question

A current carrying loop in the form of a right angle isosceles triangle $ABC$ is placed in a uniform magnetic field acting along $AB.$ If the magnetic force on the arm $BC$ is $F,$ what is the force on the arm $AC$?

• A. $-\sqrt{2}\overrightarrow{F}$
• B. $-\overrightarrow{F}$
• C. $\overrightarrow{F}$
• D. $\sqrt{2}\overrightarrow{F}$

Answer 1

The correct option is B $-\overrightarrow{F}$

If the magnetic field along AB is $\overrightarrow{B}$, the force acting on BC=$\overrightarrow{F}=BIl$

Now the length of the side AC is $\sqrt{2}l$.

Also the current reverses the direction in this section as compared to that in BC.

Hence force on AC=$\sqrt{2}l(\overrightarrow{B}\times \overrightarrow{I})$

$=-\sqrt{2}BILsin\left({45}^{\circ }\right)$

$=-BIl=-\overrightarrow{F}$