Problem 33 - University Physics with Modern Physics

Question

A straight \(2.00-\mathrm{m}, 150-\mathrm{g}\) 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 scem 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

(a) 13,364 A, impractical; (b) perpendicular to the field.

Step 1: Calculate the Gravitational Force on the Wire

The weight of the wire can be calculated using the formula for gravitational force \( F_g = m \times g \), where \( m = 150\, \text{g} = 0.150\, \text{kg} \) is the mass of the wire, and \( g = 9.8\, \text{m/s}^2 \) is the acceleration due to gravity. Hence, \[ F_g = 0.150 \times 9.8 = 1.47 \text{ N} \]

Step 2: Convert Earth's Magnetic Field to Tesla

The Earth's magnetic field is given in gauss, but we need it in tesla for the calculations. One gauss is equivalent to \( 1 \times 10^{-4} \) tesla. Therefore, \[ B = 0.55 \times 10^{-4} = 5.5 \times 10^{-5} \text{ T} \]

Step 3: Use the Magnetic Force Formula

The magnetic force \( F_m \) on a current-carrying wire in a magnetic field is given by the formula \[ F_m = I \times L \times B \times \sin(\theta) \] where \( I \) is the current, \( L = 2.00\, \text{m} \) is the length of the wire, \( B \) is the magnetic field, and \( \theta \) is the angle between the wire and the magnetic field.

Step 4: Determine the Minimum Current for Full Weight Support

To support the weight of the wire, the magnetic force \( F_m \) must equal the gravitational force \( F_g \). Therefore, \( F_m = F_g = 1.47 \text{ N} \). Assuming \( \theta = 90^\circ \) (wire perpendicular to field) so that \( \sin(\theta) = 1 \), we can solve for \( I \):\[ I \times 2.00 \times 5.5 \times 10^{-5} = 1.47 \]Solving for \( I \),\[ I = \frac{1.47}{2.00 \times 5.5 \times 10^{-5}} \approx 13,364 \text{ A} \]

Step 5: Assess Practicality of the Current

The calculated current \( 13,364 \text{ A} \) is extremely high and impractical for a simple wire, indicating that it is unlikely that such a wire could support this level of current in reality.

Step 6: Determine the Wire Orientation

For the maximum magnetic force, the wire should be oriented perpendicular to the magnetic field. This means aligning the wire such that the angle \( \theta \) with respect to the magnetic field is \( 90^\circ \), thus maximizing \( \sin(\theta) \).

Key Concepts

Current-Carrying Wire

A current-carrying wire is an essential concept that bridges the gap between electricity and magnetism. When an electric current flows through a wire, it generates a magnetic field around it. This is a fundamental principle of electromagnetism, based on Ampère's circuital law. The direction of the magnetic field created by the wire can be determined using the right-hand rule.

In the context of working with magnetic fields, the length of the wire and the magnitude of the current are critical. The wire in the original exercise carries a current through a 2-meter (m) segment. This length ensures that the wire has a substantial interaction with the magnetic field, optimizing the magnetic force experienced.

The interaction between the magnetic field and this current-carrying wire results in what we refer to as the magnetic force. This force can be calculated using the formula: \[ F_m = I \times L \times B \times \sin(\theta) \] where:

• \( I \) is the current,
• \( L \) is the length of the wire,
• \( B \) is the magnetic field strength, and
• \( \theta \) is the angle between the wire and the magnetic field.

By adjusting the current and positioning of the wire, one can control the magnitude of the magnetic force applied.

Magnetic Field Strength

Magnetic field strength, often represented by the symbol \( B \), is a measure of the magnetic influence in a region of space. It is measured in units of teslas (T), but it is common in certain contexts to use gauss, where 1 tesla equals 10,000 gauss.

In many practical scenarios such as in the original exercise, the Earth's magnetic field is considered. The given field strength was 0.55 gauss, which converts to \( 5.5 \times 10^{-5} \) tesla, demonstrating the relatively weak intensity of the Earth's field compared to many artificial magnetic fields.

The strength of the magnetic field is pivotal in determining the magnetic force acting on current-carrying wires, as seen in the force formula \( F_m = I \times L \times B \times \sin(\theta) \).

This formula highlights the linear dependence of the magnetic force on field strength \( B \). Therefore, a stronger field results in a stronger force, assuming constant current and wire length.

Therefore, understanding magnetic field strength allows us to predict and manipulate the interaction forces in electromagnetic devices and applications.

Gravitational Force

Gravitational force is the force exerted by gravity on an object due to its mass. It is a fundamental interaction that attracts objects towards the center of the Earth or any other celestial body. The gravitational force can be calculated using the formula:\[ F_g = m \times g \]where:

• \( F_g \) is the gravitational force,
• \( m \) is the mass of the object, and
• \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \) on Earth.

This force, sometimes simply referred to as weight, plays a crucial role in the analysis of forces acting on objects.

In the original exercise, the wire's gravitational force was calculated as \( 1.47 \text{ N} \) based on its mass of 150 grams (0.150 kg). This gravitational force represents the downward pull that must be counteracted by the magnetic force if the wire is to "float" or be supported solely by magnetic influence.

Balancing these forces involves leveraging the wire's magnetic force to equal the gravitational pull, illustrating how electromagnetic forces can interact with gravitational forces in physics applications.