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Chapter 15: Problem 168

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 \(a / 2\) and \(2 a\) is (A) \(\frac{1}{2}\) (B) \(\frac{1}{4}\) (C) 4 (D) 1

Short Answer

Expert verified
The ratio of the magnetic fields at \(a/2\) and \(2a\) is \(\frac{1}{2}\) (option A).

Step by step solution

01

Recall Amp猫re's Law for magnetic field

Amp猫re's Law states that the magnetic field in a wire can be calculated using the equation: \( B(r) = \frac{渭_0}{2蟺r} i \) where \(渭_0\) is the permeability of free space (\(4蟺 脳 10^{-7} Tm/A\)), \(r\) is the distance from the wire's center, and i is the current enclosed in a hypothetical loop of radius r.
02

Determine the enclosed current at a/2

The current density is uniform, so we can find the enclosed current by finding the fraction of total current enclosed in the loop of radius a/2. Let J be the current density. The current in the loop of radius a/2 is: \(i_{a/2} = J * Area_{a/2}\) As the total current is uniformly distributed, \(i = J * Area_{a}\). Dividing both equations, we get: \(\frac{i_{a/2}}{i} = \frac{Area_{a/2}}{Area_{a}}\) As the areas are circular, we can then express them as: \(\frac{i_{a/2}}{i} = \frac{蟺(a/2)^2}{蟺a^2}\) Simplifying this equation gives: \(\frac{i_{a/2}}{i} = \frac{1}{4}\) So, \(i_{a/2} = \frac{1}{4}i\).
03

Find the magnetic field at a/2

Now let's find the magnetic field at a/2: \(B(a/2) = \frac{渭_0}{2蟺(a/2)}i_{a/2}\) Substitute the current enclosed, \(i_{a/2} = \frac{1}{4}i\): \(B(a/2) = \frac{渭_0i}{8蟺}\)
04

Determine the enclosed current at 2a

As the loop of radius 2a encloses the entire wire, the enclosed current equals the total current i: \(i_{2a} = i\)
05

Find the magnetic field at 2a

Now let's find the magnetic field at 2a: \(B(2a) = \frac{渭_0}{2蟺(2a)}i\) Simplify the expression: \(B(2a) = \frac{渭_0i}{4蟺}\)
06

Calculate the ratio of magnetic fields

Now, we are to determine the ratio between the magnetic fields at a/2 and 2a: \(\frac{B(a/2)}{B(2a)} = \frac{\frac{渭_0i}{8蟺}}{\frac{渭_0i}{4蟺}}\) Cancel out \(渭_0\) and i to find the ratio: \(Ratio = \frac{1}{2}\) This matches the Option (A), making the ratio of magnetic fields at a/2 and 2a equal to \(\frac{1}{2}\). So the correct answer choice is (A).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amp猫re's Law
Amp猫re's Law is a fundamental principle in electromagnetism which relates the magnetic field around a wire to the electric current flowing through it. It's often described as one of Maxwell's equations, which are the cornerstones of classical electromagnetism. Amp猫re's Law can be summarized with the equation:

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Most popular questions from this chapter

Two infinitely long, thin, insulated, straight wires lies in the \(x-y\) plane along the \(x\) - and \(y\)-axes, respectively. Each wire carries a current \(I\), respectively, in the positive \(x\)-direction and positive \(y\)-direction. The magnetic field will be zero at all points on the straight line. (A) \(y=x\) (B) \(y=-x\) (C) \(y=x-1\) (D) \(y=-x+1\)

Two short bar magnets of magnetic moments \(M\) each are arranged at the opposite corners of a square of side \(d\) such that their centres coincide with the corners and their axes are parallel. If the like poles are in the same direction, the magnetic induction at any of the other corners of the square is (A) \(\frac{\mu_{0}}{4 \pi} \frac{M}{d^{3}}\) (B) \(\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}}\) (C) \(\frac{\mu_{0}}{4 \pi} \frac{M}{2 d^{3}}\) (D) \(\frac{\mu_{0}}{4 \pi} \frac{M^{2}}{2 d^{3}}\)

Ratio of magnetic field at the centre of a current carrying coil of radius \(R\) and at a distance of \(3 R\) on its axis is (A) \(10 \sqrt{10}\) (B) \(20 \sqrt{10}\) (C) \(2 \sqrt{10}\) (D) \(\sqrt{10}\)

The magnetic susceptibility of a material of a rod is 499. Permeability of vacuum is \(4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\). Absolute permeability of the material of the rod in henry per metre is (A) \(\pi \times 10^{-4}\) (B) \(2 \pi \times 10^{-4}\) (C) \(3 \pi \times 10^{-4}\) (D) \(4 \pi \times 10^{-4}\)

There is a small metallic ring of radius \(l_{0}\) having negligible resistance placed perpendicular to a constant magnetic field \(B_{0}\). One end of a rod is hinged at the centre of ring \(O\) and other end is placed on the ring. Now rod is rotated with constant angular velocity \(\omega_{0}\) by some external agent and circuit is connected as shown in Fig. \(15.55\), initially switch is open and capacitor is uncharged. If switch \(S\) is closed at \(t=0\), then calculate heat loss from the resistor \(R_{2}\) from \(t=0\) to the instant when voltage across the capacitor becomes \(v_{0}\) (Assume plane of ring to be horizontal and friction to be an absent at all the contacts). (Assume, \(R_{2}=2 R_{1}\), \(\left.B_{0} l_{0}^{2} \omega_{0}=4 v_{0}\right)\) (A) \(\frac{1}{2} C v_{0}^{2}\) (B) \(\frac{1}{6} C v_{0}^{2}\) (C) \(\frac{2}{3} C v_{0}^{2}\) (D) \(\frac{1}{3} C v_{0}^{2}\)
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