Let B_1 be the magnitude of the magnetic fiel | vedclass

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(C) The magnetic field at the center of a circular coil is given by $B_1 = \frac{\mu_0 I}{2R}$.The magnetic field at an axial distance $x$ from the ce

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$64 : 125$

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(C) The magnetic field at the center of a circular coil is given by $B_1 = \frac{\mu_0 I}{2R}$.The magnetic field at an axial distance $x$ from the center is given by $B_2 = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.We can express $B_2$ in terms of $B_1$ using the angle $	heta$ subtended by the radius at the point on the axis,where $\sin 	heta = \frac{R}{\sqrt{R^2 + x^2}}$.Thus,$B_2 = B_1 \sin^3 	heta$.Given $x : R = 3 : 4$,we have $x = 3k$ and $R = 4k$. The hypotenuse of the triangle formed by $x$ and $R$ is $\sqrt{R^2 + x^2} = \sqrt{(4k)^2 + (3k)^2} = 5k$.Therefore,$\sin 	heta = \frac{R}{\sqrt{R^2 + x^2}} = \frac{4k}{5k} = \frac{4}{5}$.Substituting this into the expression for the ratio:$\frac{B_2}{B_1} = \sin^3 	heta = \left(\frac{4}{5}ight)^3 = \frac{64}{125}$.

Let $B_1$ be the magnitude of the magnetic field at the center of a circular coil of radius $R$ carrying current $I$. Let $B_2$ be the magnitude of the magnetic field at an axial distance $x$ from the center. For $x : R = 3 : 4$,the ratio $\frac{B_2}{B_1}$ is:

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