SATHEE: Electro Magnetic Induction And Alternating Currents Question 446

Electro Magnetic Induction And Alternating Currents Question 446

Question: 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 $α$ . The induced emf in the loop at an instant when its side is ‘a’ is

Options:

A) $2 a α B$

B) $a^{2} α B$

C) $2 a^{2} α B$

D) $a α B$

Show Answer

Answer:

Correct Answer: A

Solution:

At any time t, the side of the square a

$= (a_{0} - α t)$ ,

where $a_{0} = s i d e$ at t=0.

At this instant, flux through the square: $ϕ = B A,$

$\cos 0^{o} = B {(a_{0} - α t)}^{2}$

$∴$ emf induced

$E = - \frac{d ϕ}{d t}$

$\Rightarrow, E = - B .2 (a_{0} - α t) (0 - α) = + 2 α a B$

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Electro Magnetic Induction And Alternating Currents Question 446

Question: 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 α . The induced emf in the loop at an instant when its side is ‘a’ is

Options:

Answer:

Solution:

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Question: 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 $α$ . The induced emf in the loop at an instant when its side is ‘a’ is