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)`.

Answer

B, C, D

Explanation

Magnetic moment of the loop `M=(iA)hatj`
`=(I_0L^2)hatk`
Magnetic field `B=(Bcos45^@)hati+(Bsin45^@)hatj`
`B/sqrt2(hati+hatj)`
a. torque acting on the loop `tau=MxxB`
`=(I_0L^2hatk)xx[B/sqrt2(hati+hatj)]`
`:. Tau=(I_0L^2B)/sqrt2 (hatj-hati)` or `|tau|=I_0L^2B`
b. Axis of rotation coincides with the torque and since torque is along `hatj-hati` direction or parallel to `QS`. Therefore, the loop will rotate about an axis passing through `Q` and `S` as shown in the figure.

Angular acceleration `alpha=|tau|/I`
where `I=`moment of inertia of loop about QS.
`I_(QS)=I_(PR)=I_(ZZ)`
(from the theorem of perpendicular axis)
But `I_(QS)=I_(PR)`
`:. 2I_(QS)=I_(ZZ)=4/3ML^2`
`I_(QS)=2/3ML^2`
`:. alpha=(|tau|)/I=(I_0L^2B)/(2/3ML^2)=3/2(I_0B)/M`
`:.` Angle by which the frame rotates in time `/_\t` is
`theta=1/2alpha(/_\t)^2`
`or theta=3/4 (I_0B)/M.(/_\t)^2`