Question

A wire abc is carrying current I. It is bent as shown in the figure and is placed in a uniform magnetic field B. Length ab = l and $\angle abc={45}^{\circ }$. What is the ratio of force on bc to that on ab?

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $1$
• D. $\frac{3}{2}$

Answer 1

The correct option is C $1$

$\frac{F_1}{F_2}=\frac{Bi{l}_{1}cos45}{Billsin{90}^{\circ }}\frac{F_1}{F_2}=1$