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 `B_0`. On right side of X'X, magnetic field strength `2B_0`Both fields are directed inside the page. The magnetic force experienced by the wire would be

Options

A. `3 IB_0R`
B. `2 IB_0R`
C. `sqrt(10) IB_0R`
D. `sqrt5 IB_0R`

Answer

c

Explanation

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

Force experienced by AB is, `F_1=IB_0(sqrt2R)`
Force experienced by BC is, `F_2=Ixx2B_0xxsqrt2R`
Resultant magnetic force acting on the wire is,
`F=sqrt(F_1^2+F_2^2)=sqrt10lB_0R`