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Question

The magnetic field due to a current carrying square loop of side a at a point located symmetrically at a distance of a/2 from its centre.
1232114_3a902d2aee164f249c2780cda76c4d46.png

A
2μ0i3πa
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B
μ0i6πa
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C
2μ0i3πa
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D
zero
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Solution

The correct option is C 2μ0i3πa
Given: Side=a ; distance=a2
Solution:
Bnet =4Bcos45=42B
Bnet =22B .........(1)
From Fig

sinα=a/23a2=13
d=a2 But ,B=μ0i4πd(sinα+sinβ)=μoi4πa/2(2sinα)
B=μ0i24πa(2/3).........(2)
Equation(2)in(1)
Bnet =/4/2×μ0i2/2/4π3
Bnet =2μ0i3πa
So,the correct option:C

2008306_1232114_ans_c34572e868ef471bb9b29a77d45b7d15.png

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