A) \[\frac{\left( {{\phi }_{2}}-{{\phi }_{1}} \right)}{R}\]
B) \[\frac{n\,\left( {{\phi }_{2}}-{{\phi }_{1}} \right)}{R}\]
C) \[\frac{\,\left( {{\phi }_{2}}-{{\phi }_{1}} \right)}{nR}\]
D) \[\frac{nR}{\,\left( {{\phi }_{2}}-{{\phi }_{1}} \right)}\]
Correct Answer: B
Solution :
Induced emf is \[|e|=n\frac{\Delta \phi }{\Delta t}\] Now, \[\Delta q=i\Delta t=\frac{e}{R}\Delta t=\frac{n\Delta \phi }{R\Delta t}\times \Delta t\] \[=\frac{n\Delta \phi }{R}=\frac{n({{\phi }_{2}}-{{\phi }_{1}})}{R}\]You need to login to perform this action.
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