1.

A uniform magnetic field vec(B) fills a cylindrical volumes of radius R. A metal rod CD of length l is placed inside the cylinder along a chord of the circular cross-section as shown in the figure. If the magnitude of magnetic field increases in the direction of field at a constant rate dB//dt, find the magnetitude and direction of the EMF induced in the rod.

Answer»


Solution :`E^(,)=E_("electrostatic")`
`E=E_("induce")`
`ointvec(E).vec(DL)=d/(dt)vec(B).vec(A)`
`E2pir=(dB)/(dt).pir^(2)`
`E=r/2 (dB)/(dt)`

`V_(D)-V_(A)=-int vec(E).dvec(x) =-intE'dx cos (pi-0)`
`=int Edx COSTHETA`
`V_(D)-V_(A)=1/2 intr(dB)/(dt)dx costheta=1/2 (dB)/(dt)int bdx`
`=1/2 (dB)/(dt).b int_(0)^(L//2) dx=1/2(dB)/(dt)sqrt(R^(2)-L^(2)/4)xxL/2`
`V_(D)-V_(C )=L/2sqrt(R^(2)-L^(2)/4)(dB)/(dt)`
`V_(induced)=A(dB)/(dt)=2xx1/2xxL/2xxsqrt(R^(2)-L^(2)/4)(dB)/(dt)`
`V_(induce)=L/2sqrt(R^(2)-L^(2)/4)(dB)/(dt)`

`V_(CO)+V_(DC)+V_(OC)=1/2 sqrt(R^(2)-L^(2)/4)(dB)/(dt)`
`V_(CO)` and `V_(OC)` are `0`, because `E` and `dl` are `_|_`
`V_(CD)=L/2 sqrt(R^(2)-L^(2)//4) (dB)/(dt)`


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