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Q.

A standing wave y = 2A sinkx cosωt is setup in the wire AB fixed at both ends by two vertical walls (see the figure). The region between the wall contains a constant magnetic field B. Now answer the following questions :The wire is found to vibrate in the 3rd harmonic. The maximum emf induced across it is nABωk. The value of n is

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answer is 4.

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Detailed Solution

λ2+λ2+λ2=AB⇒AB=3λ2=3πky=2Asin⁡(kx)cos⁡(ωt)v=∂y∂t=−2Aωsin⁡(kx)sin⁡(ωt)vmax=−2Aωsin⁡(kx)dε=Bvmaxdxε=−2BAω∫0x=3πk sin⁡(kx)dx=−2BAω-cos⁡(kx)k03πk=2BAωkcos⁡k3πk-cosk(0)|ε|=4BAωk

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