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Q.

The magnetic field due to a current carrying loop of radius 3cm at a point on its axis at the distance of 4cm from the centre is $54 μ T .$ What will be its value at the centre

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a

$75 μ T$

b

$125 μ T$

c

$150 μ T$

d

$250 μ T$

answer is A.

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Detailed Solution

$B = \frac{μ_{0} i r^{2}}{2 {(r^{2} + x^{2})}^{\frac{3}{2}}}$
$54 \times 10^{- 6} = \frac{4 π \times 10^{- 7} (i) {(3 \times 10^{- 2})}^{2}}{2 {(3^{2} + 4^{2})}^{\frac{3}{2}} {(10^{- 4})}^{\frac{3}{2}}}$
$54 \times 10^{- 6} = \frac{μ_{0} i (9 \times 10^{- 4})}{2 (125) 10^{- 6}}$
$μ_{0} i = \frac{108 \times 125 \times 10^{- 12}}{9 \times 10^{- 4}}$
At centre $= 250 \times 10^{- 6} = 250 μ T$

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