question_answer

A resistance less ring has 2 bulbs A and B rated at 2V, 19 W and 2V, 29 W respectively. The ring encloses an ideal solenoid whose magnetic field is as shown. The radius of solenoid is 1 m and the number of turns/length =1000/m. The current changes at rate of 9 A/sec Find the value of P if power dissipated in bulb B is \[180{}^\circ \] watt.       

A) 4     

B) 6            

C) 8         

D) 11    

Correct Answer: A

Solution :

[a] Resistance of bulb \[A=\frac{{{v}^{2}}}{P}=\frac{4}{10}=0.4\] Resistance of bulb \[B=\frac{{{v}^{2}}}{P}=0.2\] \[emf=\frac{d\phi }{dt}=\frac{d}{dt}({{\mu }_{0}}nI\times A)\] \[={{\mu }_{0}}n\times A\times \frac{dI}{dt}={{10}^{-7}}\times 4\pi \times 1000\times \pi {{(1)}^{2}}\times 9\] \[v=36\times {{10}^{-3}}\] \[I=\frac{v}{{{\operatorname{R}}_{eq}}}=\frac{36\times {{10}^{-3}}}{0.6}=6\times {{10}^{-2}}\,A\] Power dissipated through bulb \[B={{I}^{2}}R\] \[=36\times {{10}^{-4}}\times 0.2=7.2\times {{10}^{-4}}\,watt\]