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

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150 $μ$ T

125 $μ$ T

250 $μ$ T,

75 $μ$ T

answer is A.

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Detailed Solution

The magnetic field due to a current carrying circular coil is given by
$B = \frac{μ_{0} {nir}^{2}}{2 {(r^{2} + x^{2})}^{3 / 2}}$
or $54 = \frac{μ_{0} ni (3)^{2}}{2 (9 + 16)^{3 / 2}} = \frac{μ_{0} ni \times 9}{2 \times (5)^{2}}$
$\begin{aligned} ∴ μ_{0} ni = \frac{54 \times 250}{9} = 1500 \\ B_{centre} = \frac{μ_{0} ni}{2 r} = \frac{1500}{2 \times 3} = 250 μT \end{aligned}$