InterviewSolution
Saved Bookmarks
| 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» `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)` |
|