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Physics-

General

Easy

Question

In the given figure and magnetic field at $O$ will be

$\frac{2 μ_{0} i}{3 π a} \sqrt{4 - π^{2}}$
$\frac{μ_{0} i}{3 π a} \sqrt{4 + π^{2}}$
$\frac{2 μ_{0} i}{3 π a^{2}} \sqrt{4 + π^{2}}$
$\frac{μ_{0} i}{3 π a} \sqrt{4 - π^{2}}$

The correct answer is: $fraction numerator mu subscript 0 end subscript i over denominator 3 pi a end fraction square root of 4 plus pi to the power of 2 end exponent end root$

Magnetic field at 0 due to
Part (1) : $B subscript 1 end subscript equals 0$
Part (2) : $B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator pi blank i over denominator left parenthesis a divided by 2 right parenthesis end fraction ⨂$ [along $negative Z$ -axis]
Part (3) : $B subscript 3 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator blank i over denominator left parenthesis a divided by 2 right parenthesis end fraction left parenthesis downwards arrow right parenthesis$ [along $negative Y$ -axis]
Part (4) : $B subscript 4 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator pi i over denominator left parenthesis 3 a divided by 2 right parenthesis end fraction ⨀$ [along $plus Z$ -axis]
Part (5) : $B subscript 5 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator blank i over denominator left parenthesis 3 a divided by 2 right parenthesis end fraction left parenthesis downwards arrow right parenthesis$ [along $negative Y$ -axis]
$B subscript 2 end subscript minus B subscript 4 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator pi i over denominator a end fraction open parentheses 2 minus fraction numerator 2 over denominator 3 end fraction close parentheses equals fraction numerator mu subscript 0 end subscript i over denominator 3 a end fraction ⨂$ [along $negative Z$ -axis]
$B subscript 3 end subscript plus B subscript 5 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 1 over denominator a end fraction open parentheses 2 plus fraction numerator 2 over denominator 3 end fraction close parentheses equals fraction numerator 8 mu subscript 0 end subscript i over denominator 12 pi a end fraction left parenthesis downwards arrow right parenthesis$ [along $negative Y$ -axis]
Hence net magnetic field
$B subscript n e t end subscript equals square root of open parentheses B subscript 2 end subscript minus B subscript 4 end subscript close parentheses to the power of 2 end exponent plus open parentheses B subscript 3 end subscript plus B subscript 5 end subscript close parentheses to the power of 2 end exponent end root$
$fraction numerator mu subscript 0 end subscript i over denominator 3 pi a end fraction square root of pi to the power of 2 end exponent plus 4 end root$

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