/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 A circular current carrying loop... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 68

A circular current carrying loop of radius \(R\), carries a current \(i\). The magnetic field at a point on the axis of coil is \(\frac{1}{\sqrt{8}}\) times the value of magnetic field at the centre. Distance of point from centre is (A) \(\frac{R}{\sqrt{2}}\) (B) \(\frac{R}{\sqrt{3}}\) (C) \(R \sqrt{2}\) (D) \(R\)

Short Answer

Expert verified
The distance of the point from the center is approximately \(\boxed{\frac{R}{\sqrt{3}}}\) (Option B).

Step by step solution

01

Write down the formula for the magnetic field at a point on the axis of a circular current carrying loop.

The formula for the magnetic field (B) at a point on the axis of a circular current carrying loop is given by: \[B = \frac{\mu_0 R^2 i}{2(R^2+x^2)^{\frac{3}{2}}}\] Where: - \(B\) is the magnetic field - \(\mu_0\) is the vacuum permeability constant - \(R\) is the radius of the loop - \(i\) is the current - \(x\) is the distance from the center of the loop to the point on the axis
02

Write down the given information.

The given information is: - The magnetic field at a point on the axis of the coil is \(\frac{1}{\sqrt{8}}\) times the value of the magnetic field at the center - The magnetic field at the center can be calculated using the formula with \(x = 0\): \[B_0 = \frac{\mu_0 R^2 i}{2(R^2+0)^{\frac{3}{2}}} = \frac{\mu_0 i}{2R}\] - We need to find the distance (\(x\)) from the center of the coil to the point on the axis
03

Set up an equation using the given information.

The magnetic field at a point on the axis is \(\frac{1}{\sqrt{8}}\) times the magnetic field at the center: \[\frac{B}{\frac{1}{\sqrt{8}} \times B_0} = \frac{\mu_0 R^2 i}{2(R^2+x^2)^{\frac{3}{2}}}\]
04

Solve the equation for x.

Insert the values for the magnetic fields and then solve the equation for x: \[\frac{\frac{\mu_0 R^2 i}{2(R^2+x^2)^{\frac{3}{2}}}}{\frac{1}{\sqrt{8}} \times \frac{\mu_0 i}{2R}} = 1\] \[(R^2+x^2)^{\frac{3}{2}} = \sqrt{8}R^3\] Take the cube root on both sides: \[R^2 + x^2 = \sqrt[3]{\sqrt{8}R^3}\] \[x^2 = \sqrt[3]{\sqrt{8}R^3} - R^2 \] Now, take the square root on both sides: \[x = R\sqrt{\sqrt[3]{2}-1}\] Comparing this answer with the given options, we can conclude it is closest to Option (B), which is: \[\boxed{x \approx \frac{R}{\sqrt{3}}}\]

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