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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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Key Concepts

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

Magnetic Field
When electric current moves through a wire, it generates a magnetic field around the wire. This magnetic field is an invisible force that interacts with other magnetic fields and currents. In everyday applications, this principle helps in the functioning of devices like motors and transformers.
  • The direction of the magnetic field created depends on the direction of the current.
  • The strength of the magnetic field is influenced by the amount of current flowing through the wire.
The concept of magnetic fields is crucial in understanding how forces appear between wires carrying current, this forms the basis of many electromagnetic applications.
Biot-Savart Law
The Biot-Savart Law provides a quantitative way to calculate the magnetic field created by a current-carrying wire. It relates the magnetic field at a point in space to the current flowing through the wire and its distance from the wire. This law is particularly useful in systems where the currents have complex geometries.The Biot-Savart Law is expressed mathematically as: \[d\vec{B} = \frac{\mu_0}{4\pi} \frac{Id\vec{l}\times\vec{r}}{r^3}\]where:
  • d\vec{B} is the infinitesimally small magnetic field at a point.
  • I is the current through the wire.
  • d\vec{l} is a small segment of the wire.
  • \vec{r} is the position vector from the wire segment to the point.
  • r is the distance from the wire segment to the point.
  • \mu_0 is the permeability of free space.
This law helps us appreciate how magnetic fields are distributed around currents.
Force between current-carrying wires
Parallel wires carrying current exert forces on each other due to their magnetic fields. The direction and magnitude of the force depend on the direction of the currents. In general:
  • If the currents flow in the same direction, the wires will attract each other.
  • If the currents flow in opposite directions, the wires will repel each other.
  • The force between the wires is directly proportional to the product of the currents and inversely proportional to the distance between the wires.
The formula for the force per unit length between two long parallel wires is given by:\[\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}\]where F is the force, L is the length of the wires, I_1 and I_2 are the currents in the wires, d is the distance between the wires, and \mu_0 is the permeability of free space.
Permeability of free space
The permeability of free space, denoted by \mu_0, is a fundamental physical constant. It plays a crucial role in electromagnetic theory and is used to quantify the ability of a vacuum to support the formation of a magnetic field.The standard value is \(\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}\). This constant is foundational for calculations involving magnetic fields and forces between current-carrying wires. It appears in the Biot-Savart Law and also in Ampere's Law.
  • It helps define the ampere, which is one of the key units of electric current.
  • \(\mu_0\) appears in equations that define how magnetic force acts in free space.
Understanding \(\mu_0\) allows us to predict how magnetic fields will behave in a vacuum, which is essential in many theoretical and practical applications in physics.

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Most popular questions from this chapter

Graph of force per unit length between two long parallel current carrying conductors and the distance between them (a) Straight line (b) Parabola (c) Ellipse (d) Rectangular hyperbola

An element \(\mathrm{d} \ell^{-}=\mathrm{dx} \uparrow\) (where \(\mathrm{dx}=1 \mathrm{~cm}\) ) is placed at the origin and carries a large current \(\mathrm{I}=10 \mathrm{Amp}\). What is the mag. field on the Y-axis at a distance of \(0.5\) meter ? (a) \(2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (b) \(4 \times 10^{8} \mathrm{k} \wedge \mathrm{T}\) (c) \(-2 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (d) \(-4 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\)

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) \((1 / 2)\)

In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an ele. potential \(\mathrm{V}\) and then made to describe semicircular paths of radius \(\mathrm{r}\) using a magnetic field \(\mathrm{B}\). If \(\mathrm{V}\) and \(\mathrm{B}\) are kept constant, the ratio [(Charge on the ion) / (mass of the ion)] will be proportional to. (a) \(\left(1 / r^{2}\right)\) (b) \(r^{2}\) (c) \(\mathrm{r}\) (d) \((1 / \mathrm{r})\)

A current of a \(1 \mathrm{Amp}\) is passed through a straight wire of length 2 meter. The magnetic field at a point in air at a distance of 3 meters from either end of wire and lying on the axis of wire will be (a) \(\left(\mu_{0} / 2 \pi\right)\) (b) \(\left(\mu_{0} / 4 \pi\right)\) (c) \(\left(\mu_{0} / 8 \pi\right)\) (d) zero
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