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Chapter 13: Problem 1935

The unit of ele. current "AMPERE" is the current which when flowing through each of two parallel wires spaced 1 meter apart in vacuum and of infinite length will give rise to a force between them equal to $\mathrm{N} / \mathrm{m}$ (a) 1 (b) \(2 \times 10^{-7}\) (c) \(1 \times 10^{-2}\) (d) \(4 \pi \times 10^{-7}\)

Short Answer

Expert verified
The force between two parallel wires carrying 1 Ampere of current and spaced 1 meter apart in a vacuum is \(2 \times 10^{-7} \mathrm{N/m}\).

Step by step solution

01

Understanding the physics involved

A current flowing through a wire creates a magnetic field around the wire. If two parallel wires are carrying current in the same direction, their magnetic fields will create an attractive force between them. If the currents flow in opposite directions, the magnetic fields will create a repulsive force. In this problem, we are given that the wires carry a current of 1 Ampere.
02

Set up the Biot-Savart Law

The Biot-Savart Law describes the magnetic field around a current-carrying wire as follows: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{Id\vec{l}\times\vec{r}}{r^3} \] where: - \(d\vec{B}\) is the magnetic field at a point due to an element of the wire \(d\vec{l}\), - \(\mu_0\) is the permeability of free space, - \(I\) is the current through the wire, - \(\vec{r}\) is the position vector from the wire element to the point, - and \(r\) is the distance from the point to the wire element.
03

Calculate the magnetic field around one wire due to the other wire

For simplicity, we will only calculate the magnetic field strength at a distance one meter away from each wire. According to the Ampere's Law and Biot-Savart Law, the magnetic field strength (B) at this location will be: \[ B = \frac{\mu_0 I}{2\pi d} \] where I is the current (1 Ampere), d is the distance between the wires (1 meter), and \(\mu_0 = 4\pi \times 10^{-7} Tm/A\). Plugging in the values, \[ B = \frac{4\pi \times 10^{-7} \mathrm{Tm/A} \times 1 \mathrm{A}}{2\pi \times 1 \mathrm{m}} \] \[ B = 2 \times 10^{-7} \mathrm{T} \]
04

Calculate the force acting between the wires per unit length

The force between the wires can be calculated using the following force formula: \[ F = BIL \] where \(B\) is the magnetic field strength, \(I\) is the current, and \(L\) is the length of the wire. Since we want the force per unit length, we can divide both sides by L: \[ \frac{F}{L} = BI \] Plugging in the values, \[ \frac{F}{L} = (2 \times 10^{-7} \mathrm{T})(1 \mathrm{A}) \] \[ \frac{F}{L} = 2 \times 10^{-7} \mathrm{N/m} \] So, the correct answer is: (b) \(2 \times 10^{-7}\)

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Most popular questions from this chapter

The effective length of a magnet is \(31.4 \mathrm{~cm}\) and its pole strength is \(0.5 \mathrm{~A} \mathrm{~m}\). The magnetic moment, if it is bent in the form of a semicircle will be Amp.m \(^{2}\) (a) \(0.1\) (b) \(0.01\) (c) \(0.2\) (d) \(1.2\)

A long straight wire carrying current of \(30 \mathrm{Amp}\) is placed in an external uniform mag. field of induction \(4 \times 10^{-4}\) tesla. The mag. field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point \(2 \mathrm{~cm}\) away from the wire is tesla. (a) \(10^{-4}\) (c) \(5 \times 10^{-4}\) (b) \(3 \times 10^{-4}\) (d) \(6 \times 10^{-4}\)

If two streams of protons move parallel to each other in the same direction, then they (a) Do not exert any force on each other (b) Repel each other (c) Attract each other (d) Get rotated to be perpendicular to each other.

A domain in a ferromagnetic substance is in the form of a cube of side length \(1 \mu \mathrm{m}\). If it contains \(8 \times 10^{10}\) atoms and each atomic dipole has a dipole moment of \(9 \times 10^{-24} \mathrm{~A} \mathrm{~m}^{2}\) then magnetization of the domain is \(\mathrm{A} \mathrm{m}^{-1}\) a) \(7.2 \times 10^{5}\) (b) \(7.2 \times 10^{3}\) (c) \(7.2 \times 10^{5}\) (d) \(7.2 \times 10^{3}\)

A proton and an particle are projected with the same kinetic energy at right angles to the uniform mag. field. Which one of the following statements will be true. (a) The \(\alpha\) - particle will be bent in a circular path with a small radius that for the proton. (b) The radius of the path of the \(\alpha\) - particle will be greater than that of the proton. (c) The \(\alpha\) - particle and the proton will be bent in a circular path with the same radius. (d) The \(\alpha\) - particle and the proton will go through the field in a straight line.
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