A circular loop of radius `R` is bent along a diameter and given a shapes as shown in the figure. One of the semicircles `(KNM)` lies in the ` x-z` plane with their centres and the other one `(KLM)` in the `y-z` plane with their centres at the origin. current `I` is flowing through each of the semi circles as shown in figure.
(a) A particle of charge `q` is released at the origin with a velocity `vec(v) = -v_(0)hat(i)`. Find the instantaneous force `vec(F)` on the particle . Assume that space is gravity free.
(b) If an external uniform magnetic field `B_(0) hat(j) ` is applied , determine the force `vec(F)_(1) and vec(F)_(2)` on the semicircles `KLM and KNM` due to the field and the net force ` vec(F)` on the loop.

Explanation

`(a)` The magnetic field of semicircular arc `KLM`, at its centre `vec(B_(1))=(mu_(0)I)/(4R)(-hat(i))`

The magnetic field of semicircular arc `KLM` at its centre `vec(B_(2))=(mu_(0)I)/(4R)(hat(j))`
The resultant magnetic field `vec(B_(R))=(mu_(0)I)/(4R)(hat(j)-hat(i))`

The magnetic force on a moving charge is
`vec(F)=q(vec(v)xxvec(B))`
`=(mu_(0)qI)/(4R)(-vec(v)_(0)hat(i))xx(hat(j)xxhat(i))=(mu_(0)qI)/(4R)vec(v)_(0)(-hat(k))`

`(b)` We take a differential element on arc KLM as shown in figure `(c ),` length `dl = R theta`
In vector form,` d vec(1)=dl cos theta hat(j)-dl sin theta hat(k)`
The magnetic force on a differential element `d vec(1)`
`d vec(F)=I d vec(1) xx vec(B)=I(dl cos theta hat(j)-dl sin theta hat(k))xxB hat (j)`
`=( I dl sin theta B ) hat (i) =(IRB sin theta d theta ) ( hat (i))`
Resultant force `vec(F)_(KLM)=IRBint_(0)^(pi)sin theta d theta (hat(i))`
`=2IRBA(hat(i))`
Proceeding similar to force on arc KLM we have for KNM `d vec(1)=dl cos theta hat(i)- dl sin theta hat(k)`
where `dl=R d theta ,` we get
`dF=I( dl cos theta hat(i)-dl sintheta hat(k))xxB hat(j)`
`=I(dl cos theta hat(k)+dl sin theta hat(i))B`
`=I(Rd theta)(cos theta hat(k)+sin theta hat(i))B`
` F_(KNM)=IRB [ int _(0)^(pi)cos theta d theta hat(k)+int _(0)^(pi)sin theta d theta hat(i)]`
`=IRB[0+2 hat(i)]=2IRBhat(i)`
Thus resultant force,
`vec(F)=vec(F)_(KLM)+vec(F)_(KNM)=4IRBhat(i)`