A) 1.158 V
B) 0.58 V
C) 0.29 V
D) 5.8 V
Correct Answer: B
Solution :
Here: \[n=4000,\]\[B=0.5\times {{10}^{-4}}Wb\text{/}{{m}^{2}}\] Rate of rotation of coil \[=1800\,\,\text{rev/}\min \] \[=\frac{1800}{60}=30\,\text{rev/}\sec \] \[\omega =2\pi f=2\pi \times 30=60\pi \,rad\text{/}s,\] \[r=7\,cm=0.07\,m\] Now area of coil \[A=\pi {{r}^{2}}=\pi \times {{(0.07)}^{2}}=49\pi \times {{10}^{-4}}{{m}^{2}}\] Now the maximum energy induced is \[e=BA\,n\omega \] \[=0.5\times {{10}^{-4}}\times 49\pi \times {{10}^{-4}}\times 4000\times 60\pi \] \[=0.58\,V\]
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