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Physics-

General

Easy

Question

Figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is 2r. The value of magnetic field at the centre of the loop assuming uniform wire is

$\frac{\sqrt{2} μ_{0} i}{3 π a}$
$\frac{\sqrt{2} μ_{0} i}{3 π a}$
$\frac{\sqrt{2} μ_{0} i}{π a}$
$\frac{\sqrt{2} μ_{0} i}{π a}$

The correct answer is: $fraction numerator square root of 2 mu subscript 0 end subscript i over denominator 3 pi a end fraction$

According to question resistance of wire ADC is twice that of wire ABC. Hence current flows through ADC is half that of ABC i.e. $fraction numerator i subscript 2 end subscript over denominator i subscript 1 end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction$ . Also $i subscript 1 end subscript plus i subscript 2 end subscript equals i$ Þ $i subscript 1 end subscript equals fraction numerator 2 i over denominator 3 end fraction$ and $i subscript 2 end subscript equals fraction numerator i over denominator 3 end fraction$
Magnetic field at centre O due to wire AB and BC (part 1 and 2) $B subscript 1 end subscript equals B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 2 i subscript 1 end subscript sin invisible function application 4 5 to the power of o end exponent over denominator a divided by 2 end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 2 square root of 2 i subscript 1 end subscript over denominator a end fraction circled times$
and magnetic field at centre O due to wires AD and DC (i.e. part 3 and 4) $B subscript 3 end subscript equals B subscript 4 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 2 square root of 2 i subscript 2 end subscript over denominator a end fraction$
Also i₁ = 2i₂. So (B₁ = B₂) > (B₃ = B₄)
Hence net magnetic field at centre O
$B subscript n e t end subscript equals left parenthesis B subscript 1 end subscript plus B subscript 2 end subscript right parenthesis minus left parenthesis B subscript 3 end subscript plus B subscript 4 end subscript right parenthesis$
$equals 2 cross times fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 2 square root of 2 cross times open parentheses fraction numerator 2 over denominator 3 end fraction i close parentheses over denominator a end fraction minus fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 2 square root of 2 open parentheses fraction numerator i over denominator 3 end fraction close parentheses cross times 2 over denominator a end fraction$
$equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 4 square root of 2 i over denominator 3 a end fraction left parenthesis 2 minus 1 right parenthesis circled times equals fraction numerator square root of 2 mu subscript 0 end subscript i over denominator 3 pi a end fraction circled times$

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