question_answer

A conducting square frame of side a and a long straight wire carrying current \[I\] are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity V. The emf induced in the frame will be proportional to                                                                                                                                

A)  \[\frac{1}{{{x}^{2}}}\]                   

B)  \[\frac{1}{{{(2x-a)}^{2}}}\]         

C)  \[\frac{1}{{{(2x+a)}^{2}}}\]        

D)  \[\frac{1}{(2x+a)(2x+a)}\]

Correct Answer: D

Solution :

Potential difference across PQ is \[{{V}_{P}}-{{V}_{Q}}={{B}_{1}}(a)v=\frac{{{\mu }_{0}}I}{2\pi \left( x-\frac{a}{2} \right)}av\] Potential difference across side RS of frame is \[{{V}_{S}}-{{V}_{R}}={{B}_{2}}(a)v=\frac{{{\mu }_{0}}I}{2\pi \left( x+\frac{a}{2} \right)}av\] Hence, the net potential difference in the loop will be \[{{V}_{net}}=({{V}_{P}}-{{V}_{Q}})-({{V}_{S}}-{{V}_{R}})\] \[=\frac{{{\mu }_{0}}iav}{2\pi }\left[ \frac{1}{\left( x-\frac{a}{2} \right)}-\frac{1}{\left( x+\frac{a}{2} \right)} \right]\] \[=\frac{{{\mu }_{0}}iav}{2\pi }\left( \frac{a}{\left( x-\frac{a}{2} \right)\left( x+\frac{a}{2} \right)} \right)\] Thus \[{{V}_{net}}\propto \frac{1}{(2x-a)(2x+a)}\]