JEE Mains · Physics · STD 12 -6. Electromagnetic induction
The magnetic field in a region is given by \(\overrightarrow{ B }= B _{0}\left(\frac{ x }{ a }\right) \,\hat{ k }\). A square loop of side \(d\) is placed with its edges along the \(x\) and \(y\) axes. The loop is moved with a constant velocity \(\overrightarrow{ v }= v _{0} \hat{ i }\). The emf induced in the loop is:

- A \(\frac{ B _{0} v _{0}^{2} d }{2 a }\)
- B \(\frac{ B _{0} v _{0} d }{2 a }\)
- C \(\frac{ B _{0} v _{0} d ^{2}}{ a }\)
- D \(\frac{ B _{0} v _{0} d ^{2}}{2 a }\)
Answer & Solution
Correct Answer
(C) \(\frac{ B _{0} v _{0} d ^{2}}{ a }\)
Step-by-step Solution
Detailed explanation
\(E _{1}=\frac{ B _{0}( x + d )}{ a } v _{0} d\) \(E _{2}=\frac{ B _{0}( x )}{ a } v _{0} d\) \(E _{ net }= E _{1}- E _{2}\) \(E _{ net }=\frac{ B _{0} v _{0} d ^{2}}{ a }\)
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