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A small circular loop of conducting wire has radius a and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period T. If the mass of the loop is m then
- T = √(2πm/IB)
- T = √(2m/IB)
- T = √(πm/2IB)
- T = √(πm/IB)
Correct answer: T = √(2πm/IB)
Solution
Restoring torque about the diameter is tau = -mu*B*sin(th) ~ -(I*pi*a^2)*B*th. With moment of inertia of the ring about a diameter = m*a^2/2, angular frequency^2 = mu*B/(m*a^2/2) = 2*pi*I*B/m. Hence T = 2*pi/sqrt(2*pi*I*B/m) = sqrt(2*pi*m/(I*B)).
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