Chapter: Physics· JEE

A conducting square loop is placed in a magnetic field B with its plane perpendicular to the field. The sides of the loop start shrinking at a constant rate

A conducting square loop is placed in a magnetic field B with its plane perpendicular to the field. The sides of the loop start shrinking at a constant rate \(\alpha \) The induced emf in the loop at an instant when its side is 'a' is
A) \(2a\alpha B\)
B) \({{a}^{2}}\alpha B\)
C)\(2{{a}^{2}}\alpha B\)
D)\(a\alpha B\)

Doubtly Asked 2 Jun 2021 Student doubt physics jeemain imp electromagnetic-induction Save WhatsApp Report

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Correct solution is option A ) \(2a\alpha B\)

Explaination:

"a" at any time "t" , the side of the square a= \((a_0-\alpha_t)\) , where \(\alpha_0\) side at t=0 .

at this instant , flux through the square

⇒\(\phi = BA\cos0^o = B(a_o-\alpha_t)^2 \)

⇒\(\therefore emf \ induced \ E= -\frac{d\phi}{dt}\)

⇒\(E=-B\times 2 (a_0-\alpha_t)\)

⇒E= \(2a\alpha B\)

Doubtly · 2 Jun 2021 Report

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A conducting square loop is placed in a magnetic field B with its plane perpendicular to the field. The sides of the loop start shrinking at a constant rate

1 answer

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