Question

A wire abc is carrying current i. It is bent as shown in fig and is placed in a uniform magnetic field of magnetic induction B. Length ab = l and ∠abc = 45^o. The ratio of force on ab and on bc is

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

Answer 1

Answer: (C)

1

Answer 2

Force on portion ab of wire F₁ = Bil sin 90⁰ = Bil

Force on portion bc of wire F₂ = $B i \left( \frac { l } { \sqrt { 2 } } \right) \sin 45 ^ { \circ } = B i l$.

So $\frac { F _ { 1 } } { F _ { 2 } } = 1$.