Problem 40
Question
A wire of length, \(l\) is bent in the form of circular coil of some turns. A current, \(i\) flows through the coil. The coil is placed in a uniform magnetic field, \(B\). The maximum torque on the coil can be (a) \(\frac{i B l^{2}}{2 \pi}\) (b) \(\frac{i B I^{2}}{4 \pi}\) (c) \(\frac{i B l^{2}}{\pi}\) (d) \(\frac{2 i B l^{2}}{\pi}\)
Step-by-Step Solution
Verified Answer
The maximum torque on the coil is \( \frac{i B l^{2}}{4 \pi} \), i.e., option (b).
1Step 1: Understand the Problem
We are given a wire of length \( l \) that is bent into a circular coil with some number of turns, and a current \( i \) flows through it. The coil is placed in a uniform magnetic field \( B \). The question asks for the maximum torque on the coil.
2Step 2: Recall the Torque Formula for a Current Loop
The torque \( \tau \) on a current loop in a magnetic field is given by the formula \( \tau = n \cdot i \cdot A \cdot B \cdot \sin \theta \), where \( n \) is the number of turns, \( i \) is the current, \( A \) is the area of the coil, and \( \theta \) is the angle between the plane of the coil and the magnetic field. The maximum torque occurs when \( \sin \theta = 1 \). Therefore, \( \tau = n \cdot i \cdot A \cdot B \).
3Step 3: Express the Area of the Coil in Terms of the Length
Since the wire is bent into a circle, its circumference is equal to the length of the wire, \( l \). Thus, \( 2 \pi r = l \) where \( r \) is the radius. Solving for \( r \), we get \( r = \frac{l}{2 \pi} \). The area \( A \) of the coil is \( \pi r^2 \). Substitute \( r \) into the formula to get \( A = \pi \left(\frac{l}{2 \pi}\right)^2 = \frac{l^2}{4 \pi} \).
4Step 4: Calculate the Number of Turns
Assuming the wire is bent into one complete circle with the available length \( l \), the number of turns \( n \) is 1, given that all the length has been used for one loop.
5Step 5: Substitute Values into the Torque Formula
Substitute \( n = 1 \), \( A = \frac{l^2}{4 \pi} \) into the torque formula: \[ \tau = 1 \cdot i \cdot \frac{l^2}{4 \pi} \cdot B = \frac{i \cdot B \cdot l^2}{4 \pi}. \]
6Step 6: Identify the Correct Option
By comparing our result \( \frac{i \cdot B \cdot l^2}{4 \pi} \) with the options given:(a) \( \frac{i B l^{2}}{2 \pi} \)(b) \( \frac{i B l^{2}}{4 \pi} \)(c) \( \frac{i B l^{2}}{\pi} \)(d) \( \frac{2 i B l^{2}}{\pi} \)The correct answer is (b) \( \frac{i B l^{2}}{4 \pi} \).
Key Concepts
Current Loop in Magnetic FieldTorque FormulaCircular Coil Area Calculation
Current Loop in Magnetic Field
When a current-carrying loop is placed in a magnetic field, it experiences a torque. This is due to the interaction between the magnetic field and the current flowing through the loop. Think of the loop like a small magnetic moment. It tends to align with the external magnetic field. This behavior is at the heart of many electromagnetism applications, such as electric motors and generators.
In this scenario, the loop is circular and made from a wire of specific length bent into several turns. It's placed in a uniform magnetic field to ensure consistent interaction across its entire span. The force experienced by the current due to the magnetic field results in a turning effect known as torque.
It's important to remember that the torque will be maximized when the plane of the loop is perpendicular to the magnetic field lines, meaning the angle \( \theta \) between the field and the normal to the coil is 90 degrees. This is because torque depends on \( \sin \theta \), reaching its peak value at \( \sin 90^\circ = 1 \).
In this scenario, the loop is circular and made from a wire of specific length bent into several turns. It's placed in a uniform magnetic field to ensure consistent interaction across its entire span. The force experienced by the current due to the magnetic field results in a turning effect known as torque.
It's important to remember that the torque will be maximized when the plane of the loop is perpendicular to the magnetic field lines, meaning the angle \( \theta \) between the field and the normal to the coil is 90 degrees. This is because torque depends on \( \sin \theta \), reaching its peak value at \( \sin 90^\circ = 1 \).
Torque Formula
The formula for calculating torque \( \tau \) on a current loop in a magnetic field helps us predict how the loop responds to magnetic influences. It is given by:
This relationship is fundamental in understanding how loops of wire can convert electromagnetic energy into mechanical work. Such mechanisms are crucial in devices like galvanometers, where this principle allows for the measurement of current based on the resulting torque in the magnetic field.
- \( \tau = n \cdot i \cdot A \cdot B \cdot \sin \theta \)
- \( n \) is the number of turns of the coil
- \( i \) is the current flowing through the coil
- \( A \) is the area of a single turn of the coil
- \( B \) is the magnetic field strength
- \( \theta \) is the angle between the magnetic field and the normal to the coil
This relationship is fundamental in understanding how loops of wire can convert electromagnetic energy into mechanical work. Such mechanisms are crucial in devices like galvanometers, where this principle allows for the measurement of current based on the resulting torque in the magnetic field.
Circular Coil Area Calculation
To determine the torque on the coil, we need its area. Since the coil is circular, its geometry gives us the necessary parameters. When you have a wire that forms a circle, the area \( A \) can be found using the radius \( r \).
First, remember the relationship for the circumference of a circle: \( 2 \pi r = l \), where \( l \) is the length of the wire. By rearranging, we find \( r = \frac{l}{2\pi} \).
The area of a circle is \( A = \pi r^2 \). When we substitute \( r = \frac{l}{2\pi} \) into this area formula, we can express the area entirely in terms of \( l \):
First, remember the relationship for the circumference of a circle: \( 2 \pi r = l \), where \( l \) is the length of the wire. By rearranging, we find \( r = \frac{l}{2\pi} \).
The area of a circle is \( A = \pi r^2 \). When we substitute \( r = \frac{l}{2\pi} \) into this area formula, we can express the area entirely in terms of \( l \):
- \( A = \pi \left( \frac{l}{2\pi} \right)^2 = \frac{l^2}{4\pi} \)
Other exercises in this chapter
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