Question

A semicircular wire of radius R, carrying current I, is placed in a magnetic field as shown in fig. On left side of X'X, magnetic field strength is $B_0$, and on right side X'X, magnetic field strength is $2B_0$. Both fields are directed inside the page. The magnetic force experience by the wire would be :

• A. $3I{B}_{0}R$
• B. $2I{B}_{0}R$
• C. $\sqrt{10}I{B}_{0}R$
• D. $\sqrt{5}I{B}_{0}R$

Answer 1

The correct option is C $\sqrt{10}I{B}_{0}R$

The part AUB of the wire can be replaced by straight wire AB and BWC can be replaced by BC.

Force experienced by AB: $F_1=I{B}_0\times AB=I{B}_0\left(\sqrt{2}R\right)$

Force experienced by BC: $F_2=I\left(2B_0\right)\times BC=2I{B}_0\left(\sqrt{2}R\right)$

Resultant of $F_1$ and $F_2$ is

$F=\sqrt{F_1^2+F_2^2}=I{B}_0\left(\sqrt{2}R\right)\sqrt{1^2+2^2}=\sqrt{10}I{B}_{0}R$