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A thin wire AC shaped as a semi-circle of diameter d rotates with a constant angular frequency (omega) in a uniform magnetic field of indcution vec(B). The vector vec(omega) is parallel to vec(B) and the rotation axis XY passes through the end A of the wire and is perpendicular to the diameter AC (see Fig.). vec(B) is directed left to right in the plane of the paper. The value of the line intergral I=int vec(E)* d vec(r) taken along the wire form the point A to the point C will be (omega Bd^(2))/(?). What interger should be inserted in place of question mark. |
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Answer» Hence flux linked=`phi_(B)=B(1/2(AC)^(2)theta)` where `theta` is the angle traced in time t. By Faraday's law, the induced emf `E=oint vec(E)*dvec(L)` But we also have , `oint E*dl = e =-(d phi_(B))/(dt)` `oint E*dr = -(d)/(dt) (1/2 (d)^(2)theta)B` `=-1/2 d^(2)B((d theta)/(dt))=-(B)/(2) omega d^(2)` But we desire only the `int_(C )^(A) E*dr` which will be `1/2` or `oint E*dr`. Hence `oint_(C)^(A) E*dr = -(B)/(2) omega d^(2)`. |
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