Problem 37 - University Physics with Modern Physics

Question

A straight, 2.00 -m, \(150-\mathrm{g}\) wire carries a current in a region where the earth's magnetic field is horizontal with a magnitude of 0.55 gauss. (a) What is the minimum value of the current in this wire so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it? Does it seem likely that such a wire could support this size of current? (b) Show how the wire would have to be oriented relative to the earth's magnetic field to be supported in this way.

Step-by-Step Solution

Answer

Minimum current: 13363.64 A; orientation must be perpendicular. Such high current is unlikely safe for a normal wire.

Step 1: Understand the Problem

We have a 2.00-meter long wire of 150 grams in a magnetic field of 0.55 gauss. We need to find the current required to make the magnetic force equal to the weight of the wire, considering only gravitational force acts apart from the magnetic force.

Step 2: Convert Units

Convert the mass of the wire from grams to kilograms and the magnetic field from gauss to tesla. 1. Mass: \(150\,\text{g} = 0.150\,\text{kg}\)2. Magnetic field: \(0.55\,\text{gauss} = 0.55 \times 10^{-4}\,\text{T}\)

Step 3: Calculate the Weight of the Wire

The weight of the wire is calculated using the formula for gravitational force: \(F_g = mg\). Here, \(m = 0.150\,\text{kg}\) and \(g = 9.81\,\text{m/s}^2\).\[F_g = 0.150 \times 9.81 = 1.4715\,\text{N}\]

Step 4: Magnetic Force Formula

The magnetic force experienced by a wire is given by \(F_m = BIL \sin \theta\). To find the minimum current, we need the force to equal the weight: \(BIL \sin \theta = mg\). Assume \(\sin \theta = 1\) for maximum force.

Step 5: Solve for Current (I)

Rearrange the formula to solve for current: \[I = \frac{mg}{BL}\]Substitute \(m = 0.150\,\text{kg}\), \(g = 9.81\,\text{m/s}^2\), \(B = 0.55 \times 10^{-4}\,\text{T}\), and \(L = 2.00\,\text{m}\):\[I = \frac{0.150 \times 9.81}{0.55 \times 10^{-4} \times 2.00}\approx 13363.64\,\text{A}\]

Step 6: Orientation and Likelihood

The wire should be oriented perpendicular to the magnetic field to ensure \(\sin \theta = 1\) for maximum force. It is unlikely such a high current (approximately 13,364 Amperes) can be safely supported by a typical wire without damage due to excessive heating.

Key Concepts

Current in a Wire

Current in a wire is a fascinating concept at the heart of electricity and magnetism. In a simple sense, it refers to the flow of electric charges through a conductor. The amount of current is measured in amperes (A), and it tells us how many charges are flowing past a point in the wire per second.

Here's how current typically works in a wire:

• Electrons move through the material due to the applied voltage, creating a steady flow.
• The size of the current depends on both the voltage applied and the resistance of the wire. A higher voltage or lower resistance means more current flows.
• Ohm's Law, which is given by the formula \( I = V/R \), relates these quantities: \( I \) is current, \( V \) is voltage, and \( R \) is resistance.

Understanding the behavior of current in a wire, especially in interactions with magnetic fields, forms a critical part of electromagnetism. It helps in designing safe and efficient electrical systems.

Earth's Magnetic Field

Earth's magnetic field is an invisible force that extends from the Earth's interior out into space where it meets the solar wind. It acts much like a giant magnet with north and south poles.

• The field is horizontal at the Earth's surface and affects compasses, which we often use to find our way.
• Its strength is measured in units called gauss or tesla. For example, the typical strength of the Earth's magnetic field is about 0.55 gauss, which can also be expressed as \(0.55 \times 10^{-4}\,\text{T}\) in tesla.

This field arises from the motion of molten iron within the Earth's outer core, generating electric currents. These, in turn, produce magnetic fields. Understanding the Earth's magnetic field is crucial for navigation, as well as for understanding the protection it provides against charged particles from the sun.

Magnetic Field Orientation

Magnetic field orientation plays a critical role when we consider forces acting on a current-carrying wire in a magnetic field. The direction and orientation of this field determine how the magnetic force behaves.

• The force on the wire is strongest when it's perpendicular to the magnetic field. This is due to the relationship in the formula \( F_m = BIL \sin \theta \), where \( \theta \) is the angle between the wire and the magnetic field. The sine function (\( \sin \theta \)) reaches its maximum value of 1 when \( \theta = 90^\circ \).
• When the wire is aligned parallel or anti-parallel to the magnetic field, \( \theta = 0^\circ \) or \( 180^\circ \), the sine function returns 0, meaning no magnetic force acts in these orientations.

This concept is essential in applications ranging from electric motors to sensors, where precise control of the force exerted by magnetic fields is required. By manipulating the orientation of a wire or a current, we can change the magnitude and direction of the force exerted on it.