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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 I_{0}, 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 ML
7 years ago

Hello Student,

Please find the answer to your question

(a) As the magnetic field is in x – y plane and subtends an angles of 45° with the x –axis hence,

B_{x} = B cos 45° = B / √2 and

B_{y} = B sin 45° = B / √ 2

i.e., torque has magnitude I^{2}_{0} L^{2} 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.

I_{QS} = 1 / 2 I_{z} = 1/ 2 ( 4 / 3 ML^{2}) = 2 / 3 ML^{2}

Also τ = Iα,

∴ α = τ / I = I_{0} L^{2} B x 3 / 2ML^{2} = 3 / 2 I_{0} B / M

Here α is constant, therefore we can apply

θ = ω_{0}t + 1 / 2 αt^{2} with ω_{0} = 0, we have

θ = 1 / 2 αt^{2} = 1 / 2 (3I_{0}B / 2M) (∆t)^{2}

or θ = 3 / 4 I_{0} B / M (∆t)^{2}

Thanks

Deepak patra

askIITians Faculty

Deepak patra

askIITians Faculty

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