• # question_answer A solenoid has 2000 turns wound over a length of 0.3m, The area of its cross- section is $1.2\text{ }\times \text{ 1}{{0}^{-3}}\text{ }{{m}^{2}}.$ Around its central portion a coil of 300 turns is wound. If an initial current of 2 amp in the solenoid is reversed in 0.25 sec., the emf induced in the coil is equal to - A) $6\times {{10}^{-\,4}}\,\,V$ B) $48\,\,m\,V$ C) $6\times {{10}^{-\,2}}\,\,V$ D) $48\,\,k\,V$

Correct Answer: B

Solution :

 B due to solenoid $={{\mu }_{0}}ni$ Flux through the coil $=N\times B\times A$ Area of coil where $A=1.2\times {{10}^{-3}}{{m}^{2}}$ N= number of turns in coil = 300 Flux = $=\phi =300\,\,\times \,\,{{\mu }_{0}}ni\,\,\times \,\,1.2\,\,\times \,\,{{10}^{-\,3}}$ Current initial value = 2 Amp final current  $=-\,2Amp$
 $\Delta i=-\,4\,\,Amp$ $\Delta \phi =300{{\mu }_{0}}n\,\,\times \,\,1.2\times {{10}^{-\,3}}\Delta \,i$ $\Delta t=0.25\,\,sec$ $emf=-\frac{\Delta \phi }{\Delta t}\{Faraday\,\,law\}$$=-\,300\,\,\times \,\,{{\mu }_{0}}n\,\,\times \,\,1.2\,\,\times \,\,{{10}^{-\,3}}\frac{\Delta i}{\Delta t}$ $emf=-\,300\,\,\times \,\,4\pi \,\,\times \,\,{{10}^{-\,7}}\times \frac{2000}{0.3}\,\,\times \,\,1.2\,\,\times \,\,{{10}^{-\,3}}\frac{(-\,4)}{0.25}$$=1000\,\,\times \,\,4\pi \,\,\times \,\,{{10}^{-\,7}}\,\,\times \,\,2\,\,\times \,\,1.2\,\,\times \,\,4\,\,\times \,\,4$

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