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Physics-

General

Easy

Question

All straight wires are very long. Both AB and CD are arcs of the same circle, both subtending right angles at the centre O. Then the magnetic field at O is–

$\frac{0^{i}}{4 π R}$
$\frac{0^{i}}{4 π R} \sqrt{2}$
$\frac{0^{i}}{2 π R}$
$\frac{0^{i}}{2 π R} (π + 1)$

The correct answer is: $fraction numerator 0 to the power of i end exponent over denominator 2 pi R end fraction$

$stack B with rightwards arrow on top equals stack B with bar on top subscript text arc end text end subscript plus stack B with rightwards arrow on top subscript text sthee end text end subscript equals 0 plus open parentheses fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator I over denominator R end fraction close parentheses cross times 2 stack k with tilde on top$

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A circular current carrying loop of radius R is bent about its diameter by $.90 to the power of ring operator end exponent$ and placed in a magnetic field $stack B with rightwards arrow on top equals B subscript 0 end subscript left parenthesis stack i with bar on top plus stack j with tilde on top right parenthesis$ as shown in figure.

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The magnetic force between wires as shown in figure is :-

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The value of $fraction numerator left parenthesis 1.5 right parenthesis to the power of 3 end exponent plus left parenthesis 4.7 right parenthesis to the power of 3 end exponent plus left parenthesis 3.8 right parenthesis to the power of 3 end exponent minus 3 cross times 1.5 cross times 4.7 cross times 3.8 over denominator left parenthesis 1.5 right parenthesis to the power of 2 end exponent plus left parenthesis 4.7 right parenthesis to the power of 2 end exponent plus left parenthesis 3.8 right parenthesis to the power of 2 end exponent minus 1.5 cross times 4.7 cross times 3.8 minus 1.5 cross times 3.8 end fraction$ is -

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