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A square loop of wire consisting of a single turn is perpendicular to a uniform

ID: 1465096 • Letter: A

Question

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is 8.0 × 10-3 Wb. What is the flux that passes through the circular loop?

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is 8.0 × 10-3 Wb. What is the flux that passes through the circular loop?



Explanation / Answer

Magnetic Flux is simply
= B.A = BA cos0 = BA [Since A and B are parallel to each other hence cos 0]

Now note that B is not changing.
It is simply A that has changed from conversion of square to circle.

When Square,
B(A1) = 8 × 10^-3 Wb.............(i)

When Circle,
B(A2) = p..............................(ii) [It is p, the Magnetic flux due to circle that we have to calculate]

Divide (i) by (ii) -->

A1 / A2 = (8 × 10^-3 / p) .................(iii)

Suppose the square is made of wire of length 4s. So each side = s
So, Hence Area A1 = s^2

When the wire is moulded to cricle, its length remains unaltered
So, 2 r = 4s
=> r = 2s /
=> A2 = r^2 = 4s^2 /

Put values of A1 and A2 in (iii) -->

A1 / A2 = (8 × 10^-3 / p)

=> s^2 / (4s^2 / ) = (7.1 × e-3 / p)

=> / 4 = (7.1 × e-3 / p)

=> p = (8 × 10^-3) * (4 / )

=> p = 10.19 × 10^-3Wb

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