A uniform constant magnetic field `B` is directed at an angle of `45^(@)` to the `x axis` in the ` xy`- plane . ` PQRS` is a rigid, square wire frame carrying a steady current `I_(0)`, with its centre at the origin `O`. At time ` t = 0`, the frame is at rest in the position as shown in figure , with its sides parallel to the ` x and y` axis. Each side of the frame is of mass `M` and length `L`.
(a) What is the torque `tau` about `O` acting on the frame due to the magnetic field?
(b) Find the angle by which the frame rotates under the action of this torque in a short interval of time `Deltat`, and the axis about this rotation occurs .`( Deltat is so short that any variation in the torque during this interval may be neglected .) Given : the moment of interia of the frame about an axis through its centre perpendicular to its about an axis through its centre perpendicular to its plane is `(4)/(3) ML^(2)`.

Explanation

(a) As magnetic field is in x-y plane and subtended an angle of
`45^@` with x-axis,
`B_x=B cos 45^@=B//sqrt2`
and `B_Y=B sin 45^@ =B//sqrt2`
so, in vector form `vecB=hati(B//sqrt2)+hatj(B//sqrt2)`
and `vecM=I_0Shatk=I_0L^2hatk`
`implies vectau=vecMxxvecB =I_0L^2hatkxx (B/sqrt2hati+B/sqrt2hatj)`
`implies vectau=(I_0L^2B)/(sqrt2)xx(-hati+hatj)`
i.e,torque has magnitude `I_0L^2B` and
is directed along the line QS from
Q to S.

(b) By the theoram of perpendicular
axis, moment of inertia of the frame
about QS,
`I_(QS)=1/2I_z=1/2(4/3ML^2) =2/3ML^2`
as `tau=Ialpha implies alphatau/I =(I_0L^2Bxx3)/(2L^2M)=3/2 (I_0B)/M`
As here `alpha` is constant, equation of circular motion are
valid. Hence, from `theta=omega_0t+1/2alphat^2`, with `omega_0=0`, we have
`theta=1/2 alphat^2=1/2(3/2 (I_0B)/M)(Deltat)^2=3/4 (I_0B)/M Deltat^2`