Problem 44 - Fundamentals of Physics

Question

Figure \(28-43\) shows a wire ring of radius \(a=1.8 \mathrm{~cm}\) that is perpendicular to the general direction of a radially symmetric, diverging magnetic field. The magnetic field at the ring is everywhere of the same magnitude \(B=3.4 \mathrm{mT},\) and its direction at the ring everywhere makes an angle \(\theta=20^{\circ}\) with a normal to the plane of the ring. The twisted lead wires have no effect on the problem. Find the magnitude of the force the field exerts on the ring if the ring carries a current \(i=4.6 \mathrm{~mA}\)

Step-by-Step Solution

Answer

The force on the ring is \( 1.5 \times 10^{-8} \text{ N} \).

Step 1: Determine the Force Formula

The force exerted on a current-carrying loop in a magnetic field is given by the formula \( F = i \cdot (B \cdot A \cdot n) \), where \( i \) is the current, \( B \) is the magnetic field strength, \( A \) is the area of the loop, and \( n \) is the normal vector to the loop. The force will be influenced by the component of the magnetic field perpendicular to the loop.

Step 2: Calculate the Area of the Ring

The area \( A \) of a circle (which is the shape of the ring) is calculated using the formula \( A = \pi a^2 \). Given that the radius \( a = 1.8 \text{ cm} = 0.018 \text{ m} \), we find that:\[ A = \pi (0.018)^2 \approx 1.01788 \times 10^{-3} \text{ m}^2 \]

Step 3: Find the Perpendicular Component of the Magnetic Field

The perpendicular component of the magnetic field \( B_{\perp} \) is found using the cosine of the angle \( \theta \) between the magnetic field and the normal to the plane. Thus, \( B_{\perp} = B \cdot \cos(\theta) \). With \( B = 3.4 \text{ mT} = 3.4 \times 10^{-3} \text{ T} \) and \( \theta = 20^{\circ} \), we have:\[ B_{\perp} = 3.4 \times 10^{-3} \cdot \cos(20^{\circ}) \approx 3.2 \times 10^{-3} \text{ T} \]

Step 4: Calculate the Force on the Ring

Using the values obtained, calculate the force as:\[ F = i \cdot B_{\perp} \cdot A \]Substitute \( i = 4.6 \text{ mA} = 4.6 \times 10^{-3} \text{ A} \), \( B_{\perp} = 3.2 \times 10^{-3} \text{ T} \), and \( A = 1.01788 \times 10^{-3} \text{ m}^2 \) into the formula:\[ F = 4.6 \times 10^{-3} \times 3.2 \times 10^{-3} \times 1.01788 \times 10^{-3} \approx 1.5 \times 10^{-8} \text{ N} \]

Step 5: Present the Final Result

The final magnitude of the force exerted on the ring by the magnetic field is approximately \( 1.5 \times 10^{-8} \text{ N} \).

Key Concepts

Magnetic Field

A magnetic field is an invisible force surrounding magnetic objects, created by moving electric charges. It exerts a force on other nearby charges or magnetic materials. Picture it like an invisible web extending around a magnet. The strength of a magnetic field is measured in teslas (T) or milliteslas (mT).

• Magnitude: In our problem, the magnetic field has a uniform strength of 3.4 mT.
• Direction: The field's direction is crucial. Here, it diverges symmetrically from a central point, interacting with the loop at an angle.

Understanding the magnetic field's properties helps predict how it will interact with other objects in its vicinity.

Current-Carrying Loop

A current-carrying loop, like the wire ring in the problem, becomes a magnetic dipole when electric current flows through it. It generates its own magnetic field, which interacts with external fields.

• Shape: The loop is circular, with a radius of 1.8 cm.
• Current: The current, noted as 4.6 mA in this problem, is the flow of electric charge.

These properties influence how the loop interacts with the magnetic field surrounding it, contributing to the net force experienced by the loop.

Perpendicular Component of Magnetic Field

Not all of the magnetic field contributes to the force on the loop. Only the perpendicular component, which is at a right angle to the plane of the loop, matters. To find this:

• Direction: At the loop, the magnetic field makes an angle of 20° with the normal.
• Calculation: Using the cosine function, you can find the perpendicular component as \( B_{\perp} = B \cdot \cos(\theta) \).
• Example: Given \( B = 3.4 \text{ mT} \) and \( \theta=20^{\circ} \), \( B_{\perp} \approx 3.2 \text{ mT} \).

This calculation ensures you're using the effective field strength to determine the force.

Force Calculation

The force on a current-carrying loop in a magnetic field can be calculated with the formula \( F = i \cdot (B_{\perp} \cdot A) \). Here's how to break it down:

• Formula: The equation considers current (\( i \)), effective area (\( A \)), and the perpendicular magnetic field component (\( B_{\perp} \)).
• Area Calculation: The area \( A \) of the loop is found using \( \pi a^2 \), where \( a = 0.018 \text{ m} \). Thus, \( A \approx 1.01788 \times 10^{-3} \text{ m}^2 \).
• Substitution: With \( i = 4.6 \text{ mA} \), \( B_{\perp} = 3.2 \text{ mT} \), and \( A \), substitute into the force formula to get \( F \approx 1.5 \times 10^{-8} \text{ N} \).
• Result: This represents the magnitude of the force exerted on the loop by the field, showing how these properties and interactions culminate in a measurable force.