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

(a) `oversetrarrtau = (BI_(0)L^(2))/(sqrt2) (hati - hatj) (b) theta =(3)/(4) (BI_(0))/(M) Deltat^(2)`

Explanation

Given, magnetic field
`oversetrarrB = B(cos45^(@) hati + sin 45^(@) hatj) = (B)/sqrt2 (hati + hatj)`
`oversetrarr(A) = L^(2) hatk`
`oversetrarrtau = oversetrarrM xx oversetrarrB = i (oversetrarrA xx oversetrarrB) = hati_(0) [L^(2) hatk xx (B)/(sqrt2) (hati + hatj)]`
`= (hati_(0)L^(2B))/(sqrt2) (-hati +hatj)`
(b) It is clear from above result that axis of rotation `= ((-hati+hatj)/(sqrt2))i.e SQ`
Moment of inertia of the frame about an axis of rotation `SQ` Moment of inertia of the frame about an axis passing through `O` and perpendicular to the plane of the frame is given
`I_(0) = (4)/(3) ML^(2)`
By perpendicualr axis theorem
`2I_(SQ) =I`
`:. I_(SQ) =(I_(0))/(2) = (2)/(3) ML^(2)`
We know that `tau =I alpha`
`:. alpha = (tau)/(I) =(i_(0)BL^(2))/((2)/(3)ML^(2)) = (3)/(2) (i_(0B))/(M)`
Thus angular rotation
`theta = (1)/(2) alpha(Deltat)^(2) = (1)/(2) ((3)/(2) (i_(0)B)/(M))Deltat^(2) = (3)/(4) (i_(0)B)/(M)Deltat^(2)` .