The magnetic field due to a current-carrying | vedclass

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Question

(B) The magnetic field due to a current-carrying coil of radius $R$ on its axis at a distance $x$ from its centre is given by:$B_{	ext{axis}} = \frac

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(B) The magnetic field due to a current-carrying coil of radius $R$ on its axis at a distance $x$ from its centre is given by:$B_{	ext{axis}} = \frac{\mu_0 N I R^2}{2(x^2 + R^2)^{3/2}}$Given: $x = 8 	ext{ cm} = 8 	imes 10^{-2} 	ext{ m}$,$R = 6 	ext{ cm} = 6 	imes 10^{-2} 	ext{ m}$,and $B_{	ext{axis}} = 27 \mu 	ext{T} = 27 	imes 10^{-6} 	ext{ T}$.Substituting these values into the formula:$27 	imes 10^{-6} = \frac{\mu_0 N I (6 	imes 10^{-2})^2}{2((8 	imes 10^{-2})^2 + (6 	imes 10^{-2})^2)^{3/2}}$$27 	imes 10^{-6} = \frac{\mu_0 N I (36 	imes 10^{-4})}{2(64 	imes 10^{-4} + 36 	imes 10^{-4})^{3/2}}$$27 	imes 10^{-6} = \frac{\mu_0 N I (36 	imes 10^{-4})}{2(100 	imes 10^{-4})^{3/2}} = \frac{\mu_0 N I (36 	imes 10^{-4})}{2(10^{-2})^{3/2}}$$27 	imes 10^{-6} = \frac{\mu_0 N I (36 	imes 10^{-4})}{2 	imes 10^{-3}}$$\mu_0 N I = \frac{27 	imes 10^{-6} 	imes 2 	imes 10^{-3}}{36 	imes 10^{-4}} = \frac{54 	imes 10^{-9}}{36 	imes 10^{-4}} = 1.5 	imes 10^{-5} 	ext{ T} \cdot 	ext{m}$.Now,the magnetic field at the centre is $B_{	ext{centre}} = \frac{\mu_0 N I}{2R}$.$B_{	ext{centre}} = \frac{1.5 	imes 10^{-5}}{2 	imes 6 	imes 10^{-2}} = \frac{1.5 	imes 10^{-5}}{12 	imes 10^{-2}} = 0.125 	imes 10^{-3} 	ext{ T} = 125 \mu 	ext{T}$.

The magnetic field due to a current-carrying circular loop of radius $6 \text{ cm}$ at a point on the axis at a distance of $8 \text{ cm}$ from the centre is $27 \mu \text{T}$. The magnetic field at the centre of the current-carrying loop is: (in $\mu \text{T}$)

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