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Physics-

General

Easy

Question

The magnetic field between the plate of a capacitor when $r less than R$ is given by

$\frac{μ_{0} i_{D} r}{2 π R^{2}}$
$\frac{μ_{0} i_{D}}{2 π R}$
$\frac{μ_{0} i_{D}}{2 π r}$
Zero

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

Consider a loop of radius $r open parentheses less than R close parentheses$ between the two circular plates, placed coaxially with them. The area of the loop = $blank pi r to the power of 2 end exponent$
By symmetry magnetic field is equal in magnetic at all points on the loop. If $i subscript D end subscript superscript ´ end superscript$ is the displacement current crossing the loop and $i subscript D end subscript$ is the total displacement current between plates $i subscript D end subscript superscript ´ end superscript equals fraction numerator i subscript D end subscript r over denominator pi R to the power of 2 end exponent end fraction blank cross times pi r to the power of 2 end exponent$ . Using Ampere Maxwell’ law we have, $contour integral stack B with rightwards arrow on top bullet stack d I with rightwards arrow on top equals mu subscript 0 end subscript i subscript D end subscript superscript ´ end superscript$
or $2 pi r equals mu subscript 0 end subscript i subscript D end subscript fraction numerator pi r to the power of 2 end exponent over denominator pi R to the power of 2 end exponent end fraction o r blank B equals fraction numerator mu subscript 0 end subscript i subscript D end subscript r over denominator 2 pi R to the power of 2 end exponent end fraction$

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