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A proton, moving in negative \(y\) -direction in a magnetic field, experiences a force of magnitude \(F\), acting in the negative \(x\) -direction. a) What is the direction of the magnetic field producing this force? b) Does your answer change if the word "proton" in the statement is replaced by “electron"?

Short Answer

Expert verified
Answer: The direction of the magnetic field is in the negative \(z\)-direction. b) Would the result change if we considered an electron moving the same way as a proton? Answer: No, the result would not change. The direction of the magnetic field would still be in the negative \(z\)-direction for an electron.

Step by step solution

01

Identify given information

We are given that the force has a magnitude \(F\) and acts in the negative \(x\)-direction. Also, the proton is moving in the negative \(y\)-direction.
02

Apply the right-hand rule

From the formula mentioned earlier, \(\vec{F} = q(\vec{v} \times \vec{B})\), we can use the right-hand rule to determine the direction of the magnetic field. Point your right hand's thumb in the direction of the proton's velocity, which is the negative \(y\)-direction. Since the proton is positively charged, curl your fingers in the direction of the force, which is the negative \(x\)-direction. The direction your palm faces now will be the direction of the magnetic field.
03

Find the direction of the magnetic field

In this case, your palm will be facing downwards, which means the magnetic field is in the negative \(z\)-direction. a) So, the direction of the magnetic field is in the negative \(z\)-direction.
04

Consider the case of an electron

Now, let's see if the result changes if we replace the proton with an electron. The only difference is that an electron has a negative charge, so using the left-hand rule instead of the right-hand rule for negatively charged particles, we follow the same process as in steps 2 and 3. b) In this case, you will find that the magnetic field is still in the negative \(z\)-direction. So, the answer does not change whether we consider a proton or an electron. The direction of the magnetic field in both cases is in the negative \(z\)-direction.

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Key Concepts

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

Magnetic Force
Magnetic force is the force experienced by a charged particle when it moves through a magnetic field. This force is unique because it acts perpendicular to both the velocity of the particle and the magnetic field lines. Unlike gravitational or electrical forces, magnetic force does not do work on the particle, meaning it does not change the particle’s kinetic energy.
This kind of force can only change the direction of the particle's motion, not its speed. The magnitude of magnetic force can be calculated using the formula \( F = qvB \sin \theta \),where:
  • F is the magnetic force.
  • q is the charge of the particle.
  • v is the velocity of the particle.
  • B is the magnetic field strength.
  • \( \theta \) is the angle between the velocity vector and the magnetic field vector.
This formula highlights that the magnetic force reaches its maximum value when the particle's motion is perpendicular to the magnetic field (\( \theta = 90^\circ \)) and is zero when the particle's motion is parallel to the field.
Right-Hand Rule
The right-hand rule is a simple yet powerful tool used to determine the direction of the magnetic force on a positive charge. It involves using your right hand to visualize the interactions between the particle's velocity, the magnetic field, and the force experienced.
Here's how to use it effectively:
  • Extend your right hand.
  • Point your thumb in the direction of the particle's velocity.
  • Curl your fingers in the direction of the magnetic field lines.
  • Your palm will naturally indicate the direction of the magnetic force acting on a positive charge.
If you are dealing with a negative charge, like an electron, you would employ the left hand instead, often referred to as the left-hand rule. This adaptation accounts for the fact that the charge of the particle impacts the direction of the force due to the opposite charge sign. The right-hand rule is crucial for visualizing and solving problems about charged particles in magnetic fields.
Charged Particles in Magnetic Fields
Charged particles, such as electrons and protons, exhibit distinct behaviors when they traverse through magnetic fields. The magnetic field influences their trajectory by exerting a perpendicular force.
  • For positively charged particles like protons, the magnetic force direction is determined using the right-hand rule.
  • For negatively charged particles like electrons, the left-hand rule is used to ascertain the opposite force direction.
When these particles enter a magnetic field perpendicular to their motion, they experience a centripetal-like force which causes them to move in a circular or helical path rather than a straight line. The radius of the circular path depends on the velocity of the particle, the strength of the magnetic field, and the particle's charge and mass.
This phenomenon is utilized in many practical applications, such as in cyclotrons and mass spectrometers, where charged particles are directed and accelerated using controlled magnetic fields.

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Most popular questions from this chapter

Initially at rest, a small copper sphere with a mass of \(3.00 \cdot 10^{-6} \mathrm{~kg}\) and a charge of \(5.00 \cdot 10^{-4} \mathrm{C}\) is accelerated through a \(7000 .-\mathrm{V}\) potential difference before entering a magnetic field of magnitude \(4.00 \mathrm{~T}\), directed perpendicular to its velocity. What is the radius of curvature of the sphere's motion in the magnetic field?

The magnitude of the magnetic force on a particle with charge \(-2 e\) moving with speed \(v=1.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) is \(3.0 \cdot 10^{-18} \mathrm{~N}\). What is the magnitude of the magnetic field component perpendicular to the direction of motion of the particle?

It would be mathematically possible, for a region with zero current density, to define a scalar magnetic potential analogous to the electrostatic potential: \(V_{B}(\vec{r})=-\int_{\vec{r}_{0}}^{\vec{r}} \vec{B} \cdot d \vec{s},\) or \(\vec{B}(\vec{r})=-\nabla V_{B}(\vec{r}) .\) However, this has not been done. Explain why not.

An electron with energy equal to \(4.00 \cdot 10^{2} \mathrm{eV}\) and an electron with energy equal to \(2.00 \cdot 10^{2} \mathrm{eV}\) are trapped in a uniform magnetic field and move in circular paths in a plane perpendicular to the magnetic field. What is the ratio of the radii of their orbits?

A proton is accelerated from rest by a potential difference of \(400 .\) V. The proton enters a uniform magnetic field and follows a circular path of radius \(20.0 \mathrm{~cm} .\) Determine the magnitude of the magnetic field.

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