A wire loop carrying `I` is placed in the ` x-y`plane as shown in fig.
(a) If a particle with charge ` +Q` and mass `m` is placed at the centre `P` and given a velocity ` vec(v)` along `NP`(see figure), find its instantaneous acceleration.
( b) If an external uniform magnetic induction field ` vec(B) = Bhat(i)` is applied , find the force and the torque acting on the loop due to this field.

Explanation

(a) As in case of current-carrying straight conductor and arc,
the magnitude of B is given by
`B_1(mu_0I)/(4pid)(sinalpha+sinbeta)` and `B_2=(mu_0Iphi)/(4pir)`
So in accordance with right hand screw rule,
`(vecB_w)=mu/(4pi) 1/(a cos60^@)xx2sin 60^@(-hatk)` and
`(vecB)_(MN)=(mu_0I)/(2a)[(2pi//3)/(2pi)] hat(k)a`
and hence net `vecB` at P due to the given loop `vecB=vecB_w+vecB_A implies vecB=(mu_0)/(4pi) (2I)/a[sqrt3-pi/3](-hatk)...(i)`
Now as force on charged particle in magnetic field is
given by `vecF=q(vecvxxvecB)`
so here, `vefF=qvB sin90^@` along PF
i.e. `vecF=(mu_0)/(4pi)(2QvI)/a[sqrt3-pi/3] along PF`
and so `veca=vecF/m=10^-7(2QvI)/(ma)[sqrt3-pi/3] alongPF`
(b) As `dvecF=IdvecLxxvecB, so vecF=intIdvecLxxvecB`
As here I and `vecB` are constant
`vecF=[ointdvecL]xxvecB=0 [as ointdvecL=0]`
Further as area of coil,
`vecS=[1/3pia^2-1/2.2asin60^@xxacos 60^@] hatk=a^2[pi/3-sqrt3/4]hatk`
so `vecM=IvecS=Ia^2[pi/3-sqrt3/4]hatk`
and hence `vectau=vecMxxvecB=Ia^2B[pi/3-sqrt3/4](hatkxxhati)`
`implies vectau=Ia^2B[pi/3-sqrt3/4]hatj Nm`