# A uniform, constant magnetic field B is directed at an angle of 45 to the x axis in the xy-plane. PQRS is s rigid, square wire frame carrying a steady current I0, with its centre at the origin O. At time t = 0, the frame is at rest n the position as shown in Figure, with its sides parallel to the x and y axes. Each side of the frame is of mass M and length L.(a) What is the torque τ 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 ∆t, and the axis about this rotation occurs. (∆t is so short that any variation in the torque during this interval may be neglected.) Give : the moment of inertia of the frame about an axis through its centre perpendicular to its plane is 4/3 ML2

Deepak Patra
10 years ago
Hello Student,
(a) As the magnetic field is in x – y plane and subtends an angles of 45° with the x –axis hence,
Bx = B cos 45° = B / √2 and
By = B sin 45° = B / √ 2
i.e., torque has magnitude I20 L2 B and is directed along line QS from Q to S.
(b) According to the theorem of perpendicular axes, moment of inertia of the frame about QS.
IQS = 1 / 2 Iz = 1/ 2 ( 4 / 3 ML2) = 2 / 3 ML2
Also τ = Iα,
∴ α = τ / I = I0 L2 B x 3 / 2ML2 = 3 / 2 I0 B / M
Here α is constant, therefore we can apply
θ = ω0t + 1 / 2 αt2 with ω0 = 0, we have
θ = 1 / 2 αt2 = 1 / 2 (3I0B / 2M) (∆t)2
or θ = 3 / 4 I0 B / M (∆t)2
Thanks
Deepak patra