A) 0.009 Wb
B) 0.018 Wb
C) zero
D) 0.06 Wb
Correct Answer: B
Solution :
(b) 0.018 Wb Loop area \[=l\times b=20\times 30=600\,c{{m}^{2}}\] \[=0.06\,{{m}^{2}}\] \[B=0.3\text{ }T\] and \[\theta =90{}^\circ \] (plane is normal to field) \[\therefore \phi =BA\,\cos \,\theta =0.06\times 0.3\times \cos \,90{}^\circ \] \[=0.018\times 1=0.018\,Wb\]You need to login to perform this action.
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