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Chapter 21: Problem 4

A charge of \(-8.3 \mu \mathrm{C}\) is traveling at a speed of \(7.4 \times 10^{6} \mathrm{m} / \mathrm{s}\) in a region of space where there is a magnetic field. The angle between the velocity of the charge and the field is \(52^{\circ} .\) A force of magnitude \(5.4 \times 10^{-3} \mathrm{N}\) acts on the charge. What is the magnitude of the magnetic field?

Short Answer

Expert verified
The magnitude of the magnetic field is approximately 1.114 T.

Step by step solution

01

Understand the given situation

We need to find the magnitude of a magnetic field given a moving charge in the field, with angle between their directions provided. The force exerted by the field on the charge is given, as well as the charge's velocity and angle related to the field.
02

Recall the magnetic force formula

The magnetic force on a moving charge can be determined using the formula \( F = qvB \sin \theta \), where \( F \) is the force on the charge, \( q \) is the charge, \( v \) is the speed of the charge, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field.
03

Insert the known values into the formula

Given the values: \( F = 5.4 \times 10^{-3} \; \mathrm{N} \), \( q = -8.3 \times 10^{-6} \; \mathrm{C} \) (note that the magnitude is what's relevant here), \( v = 7.4 \times 10^6 \; \mathrm{m/s} \), and \( \theta = 52^{\circ} \). The formula becomes:\[ 5.4 \times 10^{-3} = 8.3 \times 10^{-6} \times 7.4 \times 10^6 \times B \times \sin 52^{\circ} \]
04

Solve for the magnetic field, B

Rearrange the formula to solve for \( B \):\[ B = \frac{5.4 \times 10^{-3}}{8.3 \times 10^{-6} \times 7.4 \times 10^6 \times \sin 52^{\circ}} \]Calculate the sine of \( 52^{\circ} \):\( \sin 52^{\circ} \approx 0.7880 \).Insert this into the equation and solve:\[ B \approx \frac{5.4 \times 10^{-3}}{8.3 \times 10^{-6} \times 7.4 \times 10^6 \times 0.7880} \]
05

Perform the calculation

Calculate the denominator:\[ 8.3 \times 10^{-6} \times 7.4 \times 10^6 \times 0.7880 \approx 4.8450 \]Now divide the numerator by this result:\[ B \approx \frac{5.4 \times 10^{-3}}{4.8450} \approx 1.114 \; \mathrm{T} \]

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Key Concepts

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

Magnetic Force
The concept of magnetic force is central to understanding how charges interact with magnetic fields. When a charged particle, like an electron or a proton, moves through a magnetic field, it experiences a force. This force is the magnetic force, and it's perpendicular to both the magnetic field and the direction of the charge's velocity.

Key points about magnetic force:
  • It depends on the charge (\( q \)), the velocity of the charge (\( v \)), the magnetic field (\( B \)), and the angle (\( \theta \)) between the velocity and the magnetic field.
  • The formula to calculate it is \( F = qvB \sin \theta \).
  • The direction of the force can be determined using the right-hand rule. Align your thumb in the direction of the charge's velocity, fingers in the direction of the field, and your palm will point in the direction of the force for positive charges. For negative charges, the force is in the opposite direction.
Moving Charge
A moving charge in a magnetic field experiences a unique interaction. This interaction is quantified by the magnetic force, as previously discussed. To fully understand this, let's explore what defines a moving charge and its behavior in a magnetic field.

  • A moving charge is simply a charge that doesn't stay still. When charges move, they create an electric current, which is essential in generating magnetic fields.
  • The magnetic force on the charge depends on its speed (\( v \)). Faster moving charges experience greater magnetic forces, assuming the magnetic field strength and angle remain constant.
  • Charges moving parallel or perpendicular to the field lines behave differently. If the movement is parallel, the force is zero because \( \sin 0 = 0 \), meaning no magnetic interaction occurs.
Angle Between Velocity and Field
The angle (\( \theta \)) between the velocity of a charge and the magnetic field is crucial in determining the magnetic force experienced by the charge. Understanding this angle helps predict the force鈥檚 magnitude.

  • The angle affects the term \( \sin \theta \) in the magnetic force equation. If \( \theta \) is \( 0^{\circ} \) or \( 180^{\circ} \), the force is zero because \( \sin 0 = 0 \) and \( \sin 180 = 0 \).
  • Maximum force occurs when \( \theta = 90^{\circ} \), since \( \sin 90 = 1 \).
  • This angle is regularly provided as part of problems regarding magnetic forces, and it significantly alters the outcome of calculations.
Charge Velocity
The velocity of a charge (\( v \)) moving through a magnetic field is yet another important factor in determining the magnetic force. Here's why it's so crucial:

  • Velocity is a vector quantity, meaning it has both magnitude and direction. Both aspects influence how the magnetic force is applied.
  • A higher velocity increases the magnetic force, provided all other factors remain unchanged. Since velocity is a multiplier in the formula \( F = qvB \sin \theta \), changes in speed directly impact force magnitude.
  • Understanding the direction of velocity is essential. It determines not just how strong the force will be, but also its direction when combined with the magnetic field direction.

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Most popular questions from this chapter

Multiple-Concept Example 7 discusses how problems like this one can be solved. A \(+6.00 \mu \mathrm{C}\) charge is moving with a speed of \(7.50 \times 10^{4} \mathrm{m} / \mathrm{s}\) parallel to a very long, straight wire. The wire is 5.00 \(\mathrm{cm}\) from the charge and carries a current of 67.0 \(\mathrm{A}\) in a direction opposite to that of the moving charge. Find the magnitude and direction of the force on the charge.

A copper rod of length 0.85 \(\mathrm{m}\) is lying on a frictionless table (see the drawing). Each end of the rod is attached to a fixed wire by an unstretched spring that has a spring constant of \(k=75 \mathrm{N} / \mathrm{m}\) . A magnetic field with a strength of 0.16 \(\mathrm{T}\) is oriented perpendicular to the surface of the table. (a) What must be the direction of the current in the copper rod that causes the springs to stretch? (b) If the current is \(12 \mathrm{A},\) by how much does each spring stretch?

Two parallel rods are each 0.50 m in length. They are attached at their centers to either end of a spring (spring constant \(=150 \mathrm{N} / \mathrm{m} )\) that is initially neither stretched nor compressed. When 950 \(\mathrm{A}\) of current is in each rod in the same direction, the spring is observed to be compressed by 2.0 \(\mathrm{cm}\) . Treat the rods as long, straight wires and find the separation between them when the current is present.

Particle 1 and particle 2 have masses of \(m_{1}=2.3 \times 10^{-8} \mathrm{kg}\) and \(m_{2}=5.9 \times 10^{-8} \mathrm{kg},\) but they carry the same charge \(q .\) The two particles accelerate from rest through the same electric potential difference \(V\) and enter the same magnetic field, which has a magnitude \(B\). The particles travel perpendicular to the magnetic field on circular paths. The radius of the circular path for particle 1 is \(r_{1}=12 \mathrm{cm} .\) What is the radius (in cm) of the circular path for particle 2\(?\)

ssm In New England, the horizontal component of the earth's magnetic field has a magnitude of \(1.6 \times 10^{-5}\) T. An electron is shot vertically straight up from the ground with a speed of \(2.1 \times 10^{6} \mathrm{m} / \mathrm{s}\) . What is the magnitude of the acceleration caused by the magnetic force? Ignore the gravitational force acting on the electron.
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