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Physics-

General

Easy

Question

The magnetic field at the centre of a circular coil of radius r is $pi$ times that due to a long straight wire at a distance r from it, for equal currents. Figure here shows three cases : in all cases the circular part has radius r and straight ones are infinitely long. For same current the B field at the centre P in cases 1, 2, 3 have the ratio

$(- \frac{π}{2}) : (\frac{π}{2}) : (\frac{3 π}{4} - \frac{1}{2})$
$(- \frac{π}{2} + 1) : (\frac{π}{2} + 1) : (\frac{3 π}{4} + \frac{1}{2})$
$- \frac{π}{2} : \frac{π}{2} : 3 \frac{π}{4}$
$(- \frac{π}{2} - 1) : (\frac{π}{2} - \frac{1}{4}) : (\frac{3 π}{4} + \frac{1}{2})$

The correct answer is: $open parentheses negative fraction numerator pi over denominator 2 end fraction close parentheses colon open parentheses fraction numerator pi over denominator 2 end fraction close parentheses colon open parentheses fraction numerator 3 pi over denominator 4 end fraction minus fraction numerator 1 over denominator 2 end fraction close parentheses$

Case 1 : $B subscript A end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator i over denominator r end fraction circled times$

$B subscript B end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator pi i over denominator r end fraction$ ¤
$B subscript C end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator i over denominator r end fraction$ ¤
So net magnetic field at the centre of case 1
$B subscript 1 end subscript equals B subscript B end subscript minus B subscript C end subscript minus B subscript A end subscript rightwards double arrow B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator pi i over denominator r end fraction$ ¤ ..... (i)
Case 2 : As we discussed before magnetic field at the centre O in this case
$B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator pi i over denominator r end fraction circled times$ ..... (ii)

Case 3 : $B subscript A end subscript equals 0$
$B subscript B end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator left parenthesis 2 pi minus pi divided by 2 right parenthesis i over denominator r end fraction circled times$
$B subscript C end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator i over denominator r end fraction$ ¤
$equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 3 pi i over denominator 2 r end fraction circled times$

So net magnetic field at the centre of case 3
$B subscript 3 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator i over denominator r end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction minus 1 close parentheses circled times$ ..... (iii)
From equation (i), (ii) and (iii)
$B subscript 1 end subscript colon B subscript 2 end subscript colon B subscript 3 end subscript equals pi$ ¤ : $pi circled times$ $open parentheses fraction numerator 3 pi over denominator 2 end fraction minus 1 close parentheses circled times equals negative fraction numerator pi over denominator 2 end fraction colon fraction numerator pi over denominator 2 end fraction colon open parentheses fraction numerator 3 pi over denominator 4 end fraction minus fraction numerator 1 over denominator 2 end fraction close parentheses$

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