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Chapter 13: Problem 1873

A long straight wire of radius "a" carries a steady current I the current is uniformly distributed across its cross-section. The ratio of the magnetic field at \(\mathrm{a} / 2\) and \(2 \mathrm{a}\) is (a) \((1 / 4)\) (b) 4 (c) 1 (d) \((1 / 2)\)

Short Answer

Expert verified
The ratio of the magnetic field at \(a/2\) and \(2a\) is 4. The correct answer is (b).

Step by step solution

01

Use Ampère's Law to find the magnetic field expression.

According to Ampère's Law, the magnetic field at a distance \(r\) from the axis of a long straight wire is given by the formula: \[B(r) = \frac{\mu_0 I}{2 \pi r}\] Here, \(B(r)\) is the magnetic field at a distance \(r\), \(\mu_0\) is the permeability of free space, and \(I\) is the current carried by the wire. Now we can use this formula to find the magnetic fields at the distances \(a/2\) and \(2a\).
02

Find the magnetic field at distance \(a/2\).

By substituting the distance \(a/2\) into the formula from step 1, we can find the magnetic field at this point: \[B\left(\frac{a}{2}\right) = \frac{\mu_0 I}{2 \pi \frac{a}{2}} = \frac{2\mu_0 I}{2 \pi a}\]
03

Find the magnetic field at distance \(2a\).

By substituting the distance \(2a\) into the formula from step 1, we can find the magnetic field at this point: \[B(2a) = \frac{\mu_0 I}{2 \pi 2a} = \frac{\mu_0 I}{4 \pi a}\]
04

Calculate the ratio of the magnetic fields.

Now we can compute the ratio of the magnetic fields at the two given distances: \[\frac{B\left(\frac{a}{2}\right)}{B(2a)} = \frac{\frac{2\mu_0 I}{2 \pi a}}{\frac{\mu_0 I}{4 \pi a}} = \frac{1}{1/4} = 4\] So, the ratio of the magnetic field at \(a/2\) and \(2a\) is 4. The correct answer is (b).

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Most popular questions from this chapter

If a magnet of pole strength \(\mathrm{m}\) is divided into four parts such that the length and width of each part is half that of initial one, then the pole strength of each part will be (a) \((\mathrm{m} / 4)\) (b) \((\mathrm{m} / 2)\) (c) \((\mathrm{m} / 8)\) (d) \(4 \mathrm{~m}\)

Two parallel long wires \(\mathrm{A}\) and B carry currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\). \(\left(\mathrm{I}_{2}<\mathrm{I}_{1}\right)\) when \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\) are in the same direction the mag. field at a point mid way between the wires is \(10 \mu \mathrm{T}\). If \(\mathrm{I}_{2}\) is reversed, the field becomes \(30 \mu \mathrm{T}\). The ratio \(\left(\mathrm{I}_{1} / \mathrm{I}_{2}\right)\) is (a) 1 (b) 2 (c) 3 (d) 4

Two straight long conductors \(\mathrm{AOB}\) and \(\mathrm{COD}\) are perpendicular to each other and carry currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\). The magnitude of the mag. field at a point " \(\mathrm{P}^{n}\) at a distance " \(\mathrm{a}^{\prime \prime}\) from the point "O" in a direction perpendicular to the plane \(\mathrm{ABCD}\) is (a) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}+I_{2}\right)\) (b) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}-I_{2}\right)\) (c) $\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(\mathrm{I}_{1}^{2}+\mathrm{I}_{2}^{2}\right)^{1 / 2}$ (d) $\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left[\left(I_{1} I_{2}\right) /\left(I_{1}-I_{2}\right)\right]$

A small bar magnet of moment \(\mathrm{M}\) is placed in a uniform field of \(\mathrm{H}\). If magnet makes an angle of \(30^{\circ}\) with field, the torque acting on the magnet is (a) \(\mathrm{MH}\) (b) \((\mathrm{MH} / 2)\) (c) \((\mathrm{MH} / 3)\) (d) \((\mathrm{MH} / 4)\)

A magnetic field \(B^{-}=\) Bo \(j \wedge\) exists in the region \(a
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