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The magnetic field at the center of a 50 -turn coil of radius \(15 \mathrm{~cm}\) is \(0.80 \mathrm{mT}\). Find the current in the coil.

Short Answer

Expert verified
The current in the coil is approximately 3.82 A.

Step by step solution

01

Understand the Problem

We are given a coil with 50 turns and a radius of 15 cm. The magnetic field at the center of this coil is 0.80 mT (milliTesla). We need to find out the current flowing in the coil, using the formula for the magnetic field at the center of a circular coil.
02

Identify the Formula

The magnetic field at the center of the coil, with radius \( R \) and number of turns \( N \), due to a current \( I \), is given by the formula: \[ B = \frac{\mu_0 N I}{2R} \] where \( \mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \) is the permeability of free space.
03

Convert Units

Convert the given radius from centimeters to meters for consistency in the formula. Given radius \( R = 15 \text{ cm} = 0.15 \text{ m} \).Also, convert the magnetic field from milliTeslas to Teslas for consistency. Given \( B = 0.80 \text{ mT} = 0.80 \times 10^{-3} \text{ T} \).
04

Rearrange the Formula

Rearrange the magnetic field formula to solve for the current \( I \): \[ I = \frac{2RB}{\mu_0 N} \]
05

Calculate the Current

Substitute the known values into the rearranged formula to find the current: \[ I = \frac{2 \times 0.15 \times 0.80 \times 10^{-3}}{4\pi \times 10^{-7} \times 50} \]Perform the calculations step by step to ensure accuracy.
06

Final Calculation

Complete the calculation: \[ I = \frac{2 \times 0.15 \times 0.80 \times 10^{-3}}{4\pi \times 10^{-7} \times 50} = \frac{0.24 \times 10^{-3}}{6.283 \times 10^{-5}} = \frac{0.24}{6.283} \times 10^{-2} \approx 3.82 \text{ A} \]So, the current in the coil is approximately 3.82 amperes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circular Coil Formula
The magnetic field at the center of a circular coil is a fundamental concept in electromagnetism. When a current passes through a coil of wire, it creates a magnetic field. To find the magnitude of this field at the center of the coil, we use the circular coil formula:

\[ B = \frac{\mu_0 N I}{2R} \]

This formula relates the magnetic field \( B \) to the number of turns \( N \), the current \( I \), and the radius \( R \) of the coil. The constant \( \mu_0 \) represents the permeability of free space.

When applying this formula, it is crucial to ensure that all units are consistent. This means converting radii from centimeters to meters and magnetic fields from milliTeslas to Teslas. Understanding the setup of the formula helps in solving problems related to coils and their magnetic fields effectively.
Electric Current Calculation
Calculating the electric current that flows through a coil involves rearranging the formula for the magnetic field. We do this because, in such problems, we often know the magnetic field but have to find the current. Given the magnetic field formula:

\[ B = \frac{\mu_0 N I}{2R} \]

To find the current \( I \), we rearrange the formula as follows:

\[ I = \frac{2RB}{\mu_0 N} \]

This rearrangement allows us to substitute the known values directly into the equation to solve for \( I \).

**Steps to Calculate**
  • Substitute the values for \( R \), \( B \), \( \mu_0 \), and \( N \) into the equation.
  • Ensure the units are consistent, usually in meters, Teslas, and turns.
  • Perform the arithmetic operations carefully, ensuring accuracy to solve for \( I \).
Using this method systematically ensures the right value for the electric current flowing through the coil, as seen with the result of approximately 3.82 amperes in the original problem.
Magnetic Permeability
Magnetic permeability is a property that helps us understand how materials respond to magnetic fields. In the context of our problem, we deal with the permeability of free space, represented by \( \mu_0 \). It is a constant value used in equations relating magnetic fields to currents and distances in vacuum conditions.

**Understanding \( \mu_0 \)**
  • \( \mu_0 \) is approximately equal to \( 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \).
  • This value is crucial for calculating magnetic fields created in spaces without material interference.


Having a constant like \( \mu_0 \) allows us to standardize calculations across various electromagnetic problems. It ensures that our results are accurate when materials around the coil do not alter the magnetic field. Knowing about magnetic permeability is essential for solving magnetic field problems effectively and understanding the physics underlying electromagnetic interactions.

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Most popular questions from this chapter

A positive charge moves horizontally to the right across this page and enters a magnetic field directed vertically downward in the plane of the page. (a) What is the direction of the magnetic force on the charge: (1) into the page, (2) out of the page, (3) downward in the plane of the page, or (4) upward in the plane of the page? Explain. (b) If the charge is \(0.25 \mathrm{C}\), its speed is \(2.0 \times 10^{2} \mathrm{~m} / \mathrm{s},\) and it is acted on by a force of \(20 \mathrm{~N},\) what is the magnetic field strength?

A beam of protons is accelerated to a speed of \(5.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a particle accelerator and emerges horizontally from the accelerator into a uniform magnetic field. What magnetic field (give its direction and magnitude) oriented perpendicularly to the velocity of the proton would cancel the force of gravity and keep the beam moving exactly horizontally?

A proton enters a uniform magnetic field that is at a right angle to its velocity. The field strength is \(0.80 \mathrm{~T}\) and the proton follows a circular path with a radius of \(4.6 \mathrm{~cm} .\) What are (a) the magnitude of its linear momentum and (b) its kinetic energy? (c) If its speed were doubled, what would then be the radius, momentum, and kinetic energy?

A positive charge moves horizontally to the right across this page and enters a magnetic field directed vertically downward in the plane of the page. (a) What is the direction of the magnetic force on the charge: (1) into the page, (2) out of the page, (3) downward in the plane of the page, or (4) upward in the plane of the page? Explain. (b) If the charge is \(0.25 \mathrm{C}\), its speed is \(2.0 \times 10^{2} \mathrm{~m} / \mathrm{s},\) and it is acted on by a force of \(20 \mathrm{~N},\) what is the magnetic field strength?

A straight, horizontal segment of wire carries a current in the \(+x\) -direction in a magnetic field that is directed in the \(-z\) -direction. (a) Is the magnetic force on the wire directed in the \((1)-x-,(2)+z-,(3)+y-,\) or (4) \(-y\) -direction? Explain. (b) If the wire is \(1.0 \mathrm{~m}\) long and carries a current of \(5.0 \mathrm{~A}\) and the magnitude of the magnetic field is \(0.30 \mathrm{~T}\), what is the magnitude of the force on the wire?

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