A) 2
B) 5
C) 0.4
D) 2.5
E) 0.8
Correct Answer: C
Solution :
When a wire of length\[l\]moves with a velocity \[v\] perpendicular to a magnetic field B, an induced emf is produced in the wire, given by \[e=Blv\] Given\[l=50\,cm=0.5\,m,\] \[v=300\,m\text{/}min\] \[=\frac{300}{60}=5\,m\text{/}s\] \[e=2V\] Magnetic field, \[B=\frac{e}{lv}\] \[=\frac{2}{0.5\times 5}=0.8\,T\]You need to login to perform this action.
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