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Question

Find the net magnetic filed at the centre (O) of the rectangle.


A
2μ0Iπ(b2+a2ab)
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B
2μ0Iπ(b2+a2ab2)
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C
2μ0Iπ(b2+a2a2b)
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D
2μ0Iπ(b2+a2(ab)2)
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Solution

The correct option is A 2μ0Iπ(b2+a2ab)
For the given rectangle, side AB and CD are equal in length and carries same current, hence they will produce same field at the centre of the loop. Similarly, side BC and DA will produce same field at the centre of the loop.

Let B1 be the field due to side DA.


The magnetic field due to a finite wire is,

B1=μ0I4πd(sinθ1+sinθ2)

Here, sinθ1=sinθ2=sinθ

B1=μ0I2πd(sinθ) ...(1)

Also, sinθ=b/2(a2+b2)/2

sinθ=ba2+b2

From (1), we get,

B1=μ0I2πdba2+b2=μ0I2π(a/2)ba2+b2

B1=μ0Iπaba2+b2

Similarly, magnetic field due to side AB,

B2=μ0Iπbaa2+b2

The resultant field due to all four sides is,

Bnet=2B1+2B2=2(B1+B2)

=2[μ0Iπaba2+b2+μ0Iπbaa2+b2]

=2μ0Iπa2+b2(ba+ab)

=2μ0Iπb2+a2(b2+a2ab)

=2μ0Iπ(b2+a2ab)

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