Problem 38 - College Physics

Question

A solenoid having 165 turns and a cross-sectional area of 6.75 \(\mathrm{cm}^{2}\) carries a current of 1.20 A. If it is placed in a uniform 1.12 T magnetic field, find the torque this field exerts on the solenoid if its axis is oriented (a) perpendicular to the field, (b) parallel to the field, (c) at 35. \(0^{\circ}\) with the field.

Step-by-Step Solution

Answer

(a) 0.148 N⋅m; (b) 0 N⋅m; (c) 0.085 N⋅m.

Step 1: Understand the Problem

We need to calculate the torque exerted on a solenoid placed in a magnetic field with different orientations: perpendicular, parallel, and at an angle of 35 degrees. The given parameters are: number of turns \( N = 165 \), cross-sectional area \( A = 6.75 \times 10^{-4} \ m^2 \) (converted from \( cm^2 \)), current \( I = 1.20 \ A \), and magnetic field \( B = 1.12 \ T \).

Step 2: Torque Formula for Solenoid

The formula for the torque \( \tau \) on a solenoid in a magnetic field is given by \( \tau = N \cdot I \cdot A \cdot B \cdot \sin(\theta) \), where \( \theta \) is the angle between the axis of the solenoid and the direction of the magnetic field.

Step 3: Solve for Perpendicular Orientation (90 Degrees)

When the solenoid is perpendicular to the magnetic field, \( \theta = 90^{\circ} \) and \( \sin(90^{\circ}) = 1 \). Therefore, the torque is \( \tau = 165 \cdot 1.20 \cdot 6.75 \times 10^{-4} \cdot 1.12 \cdot 1 \). Calculating this gives \( \tau = 0.148392 \ N \cdot m \).

Step 4: Solve for Parallel Orientation (0 Degrees)

When the solenoid is parallel to the magnetic field, \( \theta = 0^{\circ} \) and \( \sin(0^{\circ}) = 0 \). Hence, the torque is \( \tau = 165 \cdot 1.20 \cdot 6.75 \times 10^{-4} \cdot 1.12 \cdot 0 = 0 \ N \cdot m \).

Step 5: Solve for 35 Degrees Orientation

For an angle of \( 35^{\circ} \), the sine function yields \( \sin(35^{\circ}) \approx 0.5736 \). Therefore, the torque is \( \tau = 165 \cdot 1.20 \cdot 6.75 \times 10^{-4} \cdot 1.12 \cdot 0.5736 \). Calculating this gives \( \tau \approx 0.085145 \ N \cdot m \).

Key Concepts

Solenoid Calculations

To understand torque exerted on a solenoid, we first need to get comfortable with calculating properties of a solenoid in a magnetic field. A solenoid is a coil of wire, typically wrapped in a cylindrical shape, that produces a magnetic field when an electric current passes through it. Here are the main components necessary for solenoid calculations:

• Number of Turns ( N ): This is the total number of loops in the coil. More turns mean a stronger magnetic field.
• Cross-Sectional Area ( A ): This is the area of the coil's face, measured in square meters (or square centimeters that are converted into square meters). It affects the strength and reach of the magnetic field.
• Current ( I ): The flow of electric charge through the solenoid, measured in Amperes. More current results in a stronger magnetic field.

To move forward with solenoid calculations, we translate these parameters into the magnetic characteristics of the solenoid. The next step is to consider the solenoid's orientation within a magnetic field and how it affects the resulting torque.

Magnetic Field Orientation

The orientation of the solenoid's axis relative to the magnetic field is crucial in determining the torque experienced by the solenoid. The magnetic field's strength is denoted by B, measured in Tesla. The orientation can significantly change the effects:1. **Perpendicular Orientation (90 Degrees)**: When the solenoid's axis is perpendicular to the magnetic field, this maximizes the torque. Here, the angle \( \theta = 90^{\circ} \) and hence, \( \sin(90^{\circ}) = 1 \).
2. **Parallel Orientation (0 Degrees)**: If the solenoid's axis aligns with the magnetic field, the interaction is minimized. The angle \( \theta = 0^{\circ} \) results in \( \sin(0^{\circ}) = 0 \), meaning no torque is exerted.
3. **Oblique Orientation**: Any angle other than 0 and 90 degrees needs careful calculation. For example, at \( 35^{\circ} \), the sine value is approximately 0.5736, showing that there is some torque, but not maximum.Understanding the interaction angle helps predict the solenoid's behavior effectively and calculate how the torque will change based on its orientation.

Torque Formula Application

Applying the torque formula for a solenoid in a magnetic field involves combining the concepts of N, I, A, B, and the angle \( \theta \). The torque \( \tau \) is calculated using:\[ \tau = N \cdot I \cdot A \cdot B \cdot \sin(\theta) \]This equation reflects how each variable influences the torque:

• Number of Turns (N): More turns lead to increased torque.
• Current (I): Higher current boosts torque.
• Cross-Sectional Area (A): Larger area increases torque potential.
• Magnetic Field Strength (B): Stronger magnetic fields result in more torque.
• Angle (\( \theta \)): Torque is maximized when the solenoid is perpendicular to the field and minimized when parallel.

Calculations for different orientations (90, 0, and 35 degrees) help illustrate how each component within the formula impacts the output torque. Calculating these precisely allows for accurate predictions of how the solenoid will behave within any magnetic field setting.