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

The magnetic field at the centre P due to current in wire NM is

`B_(1)=(mu_(0))/(2pi)(I)/(r)[sin 60^(@)+sin 60^(@)]`
`B_(1)=(mu_(0))/(4pi)(I)/(a//2)[(sqrt(3))/(2)+(sqrt(3))/(2)],B_(1)=(mu_(0))/(4pi)(2Isqrt(3))/(a)`
Directed away from the reader perpendicular to the plane of paper `sin 30^(@)=(r)/(a)implies r=(a)/(2)`
`cos 30^(@)=(MS)/(a):. MS=(sqrt(3)a)/(2):.MN=sqrt(3a)`
The magnetic field at the centre P due to current in are MN is
`B_(2)=(mu_(0))/(2pi)(2pil)/(a)((theta)/(2pi))=(mu_(0))/(4pi)(2pil)/(a)[(2pi//3)/(2pi)]=(mu_(0))/(4pi)xx(2pil)/(3a)`
Directed towards the reader perpendicular to the planeof paper. The net magnetic field
`B=B_(1)-B_(2)=(mu_(0))/(4pia)(2sqrt(3l))/(a)-(mu_(0))/(4pi)(2pil)/(3a)=(mu_(0)2I)/(4pia)[sqrt(3)(pi)/(3pi)]=(mu_(0))/(4pi)xx(2I)/(a)xx(.68)`
`(` Directed away fromt he reader perpendicular to the plane of paper `)`. The force acting on the charged particle Q when it has a velocity v and is instantaneously at the centre is
`F=QvB sin theta= Q vB sin 90^(@)=Q vB`
The acceleration produced
`A=(F)/(M)=(QvB)/(m)=(Qv)/(m)[(mu_(0))/(4pi)xx(2I)/(a)xx(0.68)],A=(0.11mu_(0)IQv)/(ma)`

The direction of acceleration is given by the vector product `vec(v) xx vec(B)` or by applying Fleming's left hand rule `/_RPN=90^(@)` and `/_MPN=120^(@)`
`:. /_MPR=120-90=30^(@)`
Since `:. /_MPQ =60^(@):. /_RPQ=30^(@)`
`i.e.` the acceleration vector makes an angle of `30^(@)` with the negative `x-` axis
`(b)` The Torque acting on the loop in the magnetic field is given by `vec(tau)=vec(M)xxvec(B)`
Where `M=IA` where `A= (` Area of PMQNP `) - (` area of trianle `PMN )`
`=(1)/(3)(pia^(2))(1)/(2)xxMNxxPS=(pia^(2))/(3)(1)/(2)xxsqrt(3)axx(a)/(2)=a^(2)[(pi)/(3)(sqrt(3))/(4)]`
`vec(A)=a^(2)[(pi)/(3)-(sqrt(3))/(4)]hat(k)" ":. vec(tau)=Ia^(2)[(pi)/(3)-(sqrt(3))/(4)]hat(k)xxhat(i)B,`
`:. vec(tau)=Bla^(2)[(pi)/(3)-(sqrt(3))/(4)]hat(j)=0.614Bia^(2)hat(J)`
The force acting on the loop is zero `u=V_(0)`
`a=-(ILB)/(m)V=0` Using
`V^(2)-u^(2)=2asimpliesV_(0)^(2)=2(ILB)/(m)SimpliesS=(mV_(0)^(2))/(2ILB)`