A wire loop carrying a current `I` is placed in the `xy`-plane as shown in figure

a. If a particle chsarge +Q and mass m is placed at in centre `P` and given a velocity `v` along `NP` (see figure), find its instaneous acceleration.
If an external uniform magnetic induction field `B=Bhati` is applied, find the force and the torque acting on the loop due to this field.

Answer

(i) 0.11 (ii) 0.61

Explanation

(i) 0.11 (ii) 0.61
(i) Magnetic field at P due to arc of circle,
Subtending an angle of 120° at centre would be <

`B_(1)=1/3` (field due to circle) `=1/3((mu_(0)I)/(2a))`
`=(mu_(0)I)/(6a)` (outwards)
`=(0.16mu_(0)I)/a` (outwards)
or `B_(1)=(0.16mu_(0)I)/ahatk`
Magnetic field due to straight wire NM at P
`B_(2)=(mu_(0))/(4pi)I/r(sin 60^(@)+sin 60^(@))`
Here `r=a cos 60^(2) :. B_(2)=(mu_(0))/(4pi)I/(a cos 60^(@))(2 sin 60^(@))`
or `B_(2)=(mu_(0))/(2pi)I/a tan 60^(@)=(0.27mu_(0)I)/a` (inwards)
or `B_(2)=-(0.27mu_(0)I)/a hatk`
Now, velocity of particle can be written as
`v=v cos 60^(@)hatj=v/2hati+(sqrt(3))/2hatj`
Magnetic force
`F_(m)=q(vxxB)=(0.11mu_(0)IQV)/(2a)hatj-(0.11sqrt(3)mu_(0)IQV)/(2a)hati`
`:.` intantaneosu acceleration
`a=(F_(m))/m=(0.11mu_(0)IQv)/(2am)(hatj-sqrt(3)hati)`
(ii) In uniform magnetic field, force on a current loop is zero. Further, magnetic dipole moment of the loop will be,
`M=(IA)hatk`
Here A is the area of the loop
`A=1/3(pia^(2))-1/2[2xxa sin 60^(@)][a cos 60^(@)]`
`=(pia^(2))/3-(a^(2))/2sin120^(@)=0.61a^(2)`
`:.M=(0.61Ia^(2))hatk`
Given `B=Bhati`
`:.tau=MxxB-(0.641Ia^(2)B)hatj`