A) 0.321
B) 0.471
C) 0.235
D) 0.512
Correct Answer: B
Solution :
Given: \[\ell =1m,B=5\times {{10}^{-3}}\text{Wb/}{{\text{m}}^{\text{2}}}\] \[f=\frac{1800}{60}=30\]rotations/sec In one rotation, the moving rod of the metal traces a circle of radius \[r=\ell \] \[\therefore \]Area swept in one rotation\[=\pi {{r}^{2}}\] \[\frac{d\phi }{dt}=\frac{d}{dt}(BA)=B.\frac{dA}{dt}=\frac{B\pi {{r}^{2}}}{T}\] \[=Bf\pi {{r}^{2}}=(5\times {{10}^{-3}})\times 3.14\times 30\times 1=0.471\,\text{V}\] \[\therefore \] e.m.f. induced in a metal rod = 0.471 V
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