Q. The magnetic field $B$ due to a current-carrying circular loop of the radius $12 \, cm$ at its centre is $0.5\times 10^{- 4} \, T$ . The magnetic field due to this loop at a point on the axis at a distance of $5 \, cm$ from the centre -

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A

$3.9 \, \times \, 10^{- 5} \, T$

B

$5.2 \, \times \, 10^{- 5} \, T$

C

$2.1 \, \times \, 10^{- 5} \, T$

D

$9 \, \times \, 10^{- 5} \, T$

Solution:

$B_{0} \, =\frac{\mu _{0} I}{2 a}$
At axial point
$B=\frac{\left(\mu \right)_{0} I a^{2}}{2 \, \left(\right. a^{2} + x^{2} \left(\left.\right)^{3 / 2}}$
$\frac{B}{B_{0}}=\frac{a^{3}}{\left(\right. a^{2} + x^{2} \left(\left.\right)^{3 / 2}}$
$\Rightarrow \, B= \, B_{0} \, \frac{a^{3}}{\left(\right. a^{2} + x^{2} \left(\left.\right)^{3 / 2}}$
$= \, 0.5 \, \times \, \left(10\right)^{- 4} \, \times \frac{\left(\right. 12 c m \left(\left.\right)^{3}}{\left(\right. 144 c m^{2} + 25 c m^{2} \left(\left.\right)^{3 / 2}}$
$= \, 3.9 \, \times \, 10^{- 5} \, T.$

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