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There is small circular coil which is suspended through a wire of length l to a rigid support. Initially, thread is kept vertical and uniform magnetic field B is applied perpendicular to the plane of circular coil. Coil is slightly displaced through an angle theta_(0) and released in such a manner that while oscillating plane of the coil always remains along the thread and it does not rotate about it. Calculate emf induced in the coil as a function of time assuming t = 0 at some instant when thread is vertical. |
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Answer» Solution :Initially when thread is vertical then angle between magnetic field and area vector of coil is zero. When thread makes an angle `theta` with the vertical then angle between magnetic field and area vector of coil also becomes `theta`. Hence magnetic flux linked with the coil can be WRITTEN as follows: `phi=BA COS theta` We can apply Faraday.s law to get emf induced in the coil as follows: `epsilon=-(dphi)/(dt)=BA sin theta (dphi)/(dt)` Here angle `theta` is given to be very SMALL. Hence we use `sin theta approx theta` to rewrite the above relation as follows: `epsilon=BAtheta(dtheta)/(dt)` We know that coil PERFORMS SHM. And the angle the thread makes with the vertical can be written as follows: `theta= theta_(0) sin omegat` where `omega=sqrt((g)/(l))` On differentiating equation (ii) we get the following: `(dtheta)/(dt)=theta_(0) omega cos omegat` ...(iii) Substituting from equation (ii) and (iii) in equation (i) we get the following: `epsilon=BA(theta_(0)sin omegat)(theta_(0) omega cos omegat)` `implies epsilon =BA omega theta_(0)^(2)sin omegat cos omegat` `implies epsilon=(1)/(2)BA omega theta_(0)^(2) sin2omegat` |
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