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A magnetic field has a magnitude of \(1.2 \times 10^{-3} \mathrm{~T}\), and an electric field has a magnitude of \(4.6 \times 10^{3} \mathrm{~N} / \mathrm{C}\). Both fields point in the same direction. A positive \(1.8-\mu \mathrm{C}\) charge moves at a speed of \(3.1 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.

Short Answer

Expert verified
The net force on the charge is approximately 10.65 N.

Step by step solution

01

Understand the Forces Involved

The problem involves two kinds of forces: Electric force and Magnetic force. The net force is the vector sum of these two forces acting on the charge.
02

Calculate the Electric Force

The electric force (\( F_e \) ) acting on a charge is given by the formula: \( F_e = qE \) where \( q \) is the charge and \( E \) is the electric field strength. Substitute the given values: charge \( q = 1.8 \times 10^{-6} \mathrm{~C} \), and electric field \( E = 4.6 \times 10^3 \mathrm{~N/C} \). This results in: \[ F_e = 1.8 \times 10^{-6} \cdot 4.6 \times 10^3 = 8.28 \mathrm{~N} \]
03

Calculate the Magnetic Force

The magnetic force (\( F_m \) ) acting on a moving charge in a magnetic field is given by: \( F_m = qvB \) where \( v \) is the velocity of the charge and \( B \) is the magnetic field strength. Using the given values: charge \( q = 1.8 \times 10^{-6} \mathrm{~C} \), velocity \( v = 3.1 \times 10^6 \mathrm{~m/s} \), and magnetic field \( B = 1.2 \times 10^{-3} \mathrm{~T} \), compute:\[ F_m = 1.8 \times 10^{-6} \cdot 3.1 \times 10^6 \cdot 1.2 \times 10^{-3} = 6.696 \mathrm{~N} \]
04

Calculate the Net Force

Since the charge is moving perpendicular to both fields and the fields are in the same direction, the electric and magnetic forces are perpendicular, hence use the Pythagorean theorem to find the net force: \( F_{net} = \sqrt{F_e^2 + F_m^2} \). Substitute the forces calculated above: \[ F_{net} = \sqrt{8.28^2 + 6.696^2} = \sqrt{68.5584 + 44.8248} = \sqrt{113.3832} \approx 10.65 \mathrm{~N} \]

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Key Concepts

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

Electric Force
Electric force, a fundamental concept in electromagnetism, acts on a charged particle when it is placed in an electric field. Imagine a field similar to gravity, but instead of pulling everything down, this electric field applies a force depending on whether the charge is positive or negative.

For a positive charge, the electric force direction aligns with the electric field's direction, like in this exercise where both the field and the force are in the same line. Conversely, a negative charge would move in the opposite direction. The strength or magnitude of this force depends on both the charge amount and the field strength itself.
  • The formula used to determine electric force is:
    \( F_e = qE \)
  • Where \( q \) represents the charge and \( E \) the magnitude of the electric field.
  • In the exercise, the electric force is calculated to be \( 8.28 \text{~N} \), using the charge \( 1.8 \times 10^{-6} \text{~C} \) and electric field \( 4.6 \times 10^{3} \text{~N/C} \).
Understanding these principles helps clarify why and how forces act on charges within electric fields, which is crucial for solving many physics problems.
Magnetic Force
Magnetic force plays a pivotal role when a charged particle moves through a magnetic field. In this context, the force generated is perpendicular to both the direction of the magnetic field and that of the moving charge. This perpendicular relationship stems from the nature of magnetic forces, following the right-hand rule often mentioned in physics.

In practical terms,
  • The magnetic force formula is expressed as:
    \( F_m = qvB \)
  • Where \( q \) is the charge, \( v \) is the velocity of the particle, and \( B \) is the strength of the magnetic field.
During the exercise, the magnetic force was computed to be \( 6.696 \text{~N} \), using the given values of \( q = 1.8 \times 10^{-6} \text{ C} \), \( v = 3.1 \times 10^{6} \text{ m/s} \), and \( B = 1.2 \times 10^{-3} \text{ T} \).

This demonstrates how magnetic fields influence moving charges, resulting in forces that are crucial for applications like electric motors and magnetic resonance imaging.
Net Force
The net force experienced by a charged particle moving through electric and magnetic fields is a combination of the electric force and the magnetic force. Since both forces act perpendicular to each other in our exercise scenario, the net force is calculated using the Pythagorean theorem, which applies to right-angle vector addition.

Simply put, the net force determination requires that:
  • We use the formula:
    \( F_{net} = \sqrt{F_e^2 + F_m^2} \)
  • Where \( F_e \) and \( F_m \) are the magnitudes of the electric and magnetic forces, respectively.
In this case, with \( F_e = 8.28 \text{~N} \) and \( F_m = 6.696 \text{~N} \), this gives us the total force \( F_{net} \approx 10.65 \text{~N} \).
This final calculation shows how two independent forces can combine to create a resultant force that dictates the movement of the charge, an important consideration in the analysis and design of systems involving electromagnetic forces.

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Most popular questions from this chapter

A long solenoid has 1400 turns per meter of length, and it carries a current of \(3.5 \mathrm{~A}\). A small circular coil of wire is placed in side the solenoid with the normal to the coil oriented at an angle of \(90.0^{\circ}\) with respect to the axis of the solenoid. The coil consists of 50 turns, has an area of \(1.2 \times 10^{-3} \mathrm{~m}^{2},\) and carries a current of \(0.50 \mathrm{~A}\). Find the torque exerted on the coil.

Due to friction with the air, an airplane has acquired a net charge of \(1.70 \times 10^{-5} \mathrm{C}\) The airplane travels with a speed of \(2.80 \times 10^{2} \mathrm{~m} / \mathrm{s}\) at an angle \(\theta\) with respect to the earth's magnetic field, the magnitude of which is \(5.00 \times 10^{-5} \mathrm{~T}\). The magnetic force on the airplane has a magnitude of \(2.30 \times 10^{-7} \mathrm{~N}\). Find the angle \(\theta\) (There are two possible angles.)

An electron is moving through a magnetic field whose magnitude is \(8.70 \times 10^{-4} \mathrm{~T}\). The electron experiences only a magnetic force and has an acceleration of magnitude \(3.50 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}\). At a certain instant, it has a speed of \(6.80 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Determine the angle \(\theta\) (less than \(90^{\circ}\) ) between the electron's velocity and the magnetic field.

The two conducting rails in the drawing are tilted upward so they each make an angle of \(30.0^{\circ}\) with respect to the ground. The vertical magnetic field has a magnitude of \(0.050 \mathrm{~T}\) The \(0.20-\mathrm{kg}\) aluminum rod (length \(=1.6 \mathrm{~m}\) ) slides without friction down the rails at a constant velocity. How much current flows through the bar?

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole human bodies has a field strength of \(7.0 \mathrm{~T},\) and the current in the solenoid is \(2.0 \times 10^{2} \mathrm{~A} .\) What is the number of turns per meter of length of the solenoid? Note that the solenoid used to produce the magnetic field in this type of system has a length that is not very long compared to its diameter. Because of this and other design considerations, your answer will be only an approximation.

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