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

(i) 1 (ii) 3/4

Explanation

(i) 1 (iI) 3/4
Magnetic moment of the loop `M=(iA)hatK=(I_(0)L^(2))hatk`
Magnetic field `B=(B cos 45^(@))hati+(B sin 45^(@))hatj=B/(sqrt(2))(hati+hatj)`
(i) Torque acting on the loop `tau=MxxB`
`=(I_(0)L^(2)hatk)xx[B/(sqrt(2))(hati+hatj)] " "tau=(I_(0)L^(2)B)/(sqrt(2))(hatu-hati)` or `|tau|=I_(0)L^(2)B`
(ii) Axis of rotation coincides with the torque and since torque is in `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 theorem of perpendicular axis)
But `I_(QS)=I_(PR)`
`:.2I_(QS)=I_(ZZ)=4/3ML^(2)`
`I_(QS)=2/3ML^(2):. aklpha=(|tau|)/I=(I_(0)L^(2)B)/(2//3ML^(2))=3/2 (I_(0)B)/M`
`:.` Angle by which the frame rotates in time `Deltat` is
`theta=1/2 alpha(Deltat)^(2)` or `theta=3/4 (I_(0)B)/M . (Delta t)^(2)`