Question

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

$1 over straight x squared$
$fraction numerator 1 over denominator left parenthesis 2 straight x minus straight a right parenthesis squared end fraction$
$fraction numerator 1 over denominator left parenthesis 2 straight x plus straight a right parenthesis squared end fraction$
$fraction numerator 1 over denominator left parenthesis 2 straight x minus straight a right parenthesis left parenthesis 2 straight x plus straight a right parenthesis end fraction$

Solution

fraction numerator 1 over denominator left parenthesis 2 straight x minus straight a right parenthesis left parenthesis 2 straight x plus straight a right parenthesis end fraction

Potential difference across PQ is
$straight V subscript straight P space minus straight V subscript straight Q space equals space straight B subscript 1 space left parenthesis straight a right parenthesis straight v space equals space fraction numerator straight mu subscript straight o straight I over denominator 2 straight pi open parentheses straight x minus begin display style straight a over 2 end style close parentheses end fraction av Potential space difference space across space side space RS space of space frame space is space straight V subscript straight s space minus straight V subscript straight R space equals space straight B subscript 2 space left parenthesis straight a right parenthesis straight v space equals space fraction numerator straight mu subscript straight o straight I over denominator 2 straight pi open parentheses straight x plus begin display style straight a over 2 end style close parentheses end fraction av Hence comma space the space net space potential space difference space in space the space lop space will space be space straight V subscript net space left parenthesis straight V subscript straight P minus end subscript straight V subscript straight Q right parenthesis space minus space left parenthesis straight V subscript straight s minus space straight V subscript straight R right parenthesis equals space fraction numerator straight mu subscript straight o iav over denominator 2 straight pi end fraction space open square brackets fraction numerator 1 over denominator open parentheses straight x minus begin display style straight a over 2 end style close parentheses end fraction minus fraction numerator 1 over denominator open parentheses straight x plus begin display style straight a over 2 end style close parentheses end fraction close square brackets equals space fraction numerator straight mu subscript straight o space iav over denominator 2 straight pi end fraction space open parentheses fraction numerator straight a over denominator open parentheses straight x minus straight a over 2 close parentheses open parentheses straight x plus straight a over 2 close parentheses end fraction close parentheses straight v subscript net space proportional to space fraction numerator 1 over denominator left parenthesis 2 straight x minus straight a right parenthesis left parenthesis 2 straight x plus straight a right parenthesis end fraction$

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Electric Charges And Fields

Some More Questions From Electric Charges and Fields Chapter

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
b) Explain why two field lines never cross each other at any point?

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