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A charged particle moves along a circle under the action of possible constant electric and magnetic fields. Which of the following are possible? a. \(E=0, B=0\) \(\mathbf{b} E=0, B \neq 0\) c. \(E \neq 0, B=0\) d. \(E \neq 0, B \neq 0\)

Short Answer

Expert verified
Cases b and d are possible.

Step by step solution

01

Understanding the Particle Motion

For a charged particle to move along a circular path, there must be a centripetal force acting on it to continually change its direction of motion.
02

Analyzing the Case 1 (a. E=0, B=0)

If both the electric field (E) and the magnetic field (B) are zero, there are no forces acting on the charged particle. Without any forces, the particle cannot move in a circle. Thus, this case is impossible.
03

Analyzing the Case 2 (b. E=0, B≠0)

In this scenario, only the magnetic field is present. A magnetic field can exert a force on a moving charged particle, which is perpendicular to the direction of motion and the field. This force can provide the necessary centripetal force to maintain circular motion. Thus, this case is possible.
04

Analyzing the Case 3 (c. E≠0, B=0)

An electric field exerts a force on a charged particle in the direction of the field. This force would result in linear, not circular, acceleration as it is directed along the field lines rather than centripetally. Therefore, this case is impossible for circular motion.
05

Analyzing the Case 4 (d. E≠0, B≠0)

With both electric and magnetic fields present, these fields can interact with the charged particle. The magnetic field can provide the centripetal force for circular motion, even in the presence of an electric field. Thus, this case is possible.

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Key Concepts

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

Centripetal Force
In physics, centripetal force is essential for an object to move in a circular path. This force always acts perpendicular to the object's velocity and towards the center of the circle. Without it, an object would continue moving in a straight line due to inertia.
For a particle in circular motion, centripetal force is crucial because it constantly pulls the particle toward the circle's center, allowing it to stay on its circular path.
Mathematically, centripetal force is expressed as:
  • \( F_{c} = \frac{mv^2}{r} \)
where:
  • \( F_{c} \) is the centripetal force,
  • \( m \) is the mass of the particle,
  • \( v \) is the velocity of the particle, and
  • \( r \) is the radius of the circle.
In the case of charged particles in electromagnetic fields, a magnetic force can act as the centripetal force. This is essential for understanding how charged particles can maintain circular motion, even in the presence of electrical fields, as can be seen in scenarios where both magnetic and electric fields play a role.
Magnetic Force
A magnetic force arises from the motion of charged particles in a magnetic field. If you have a charged particle moving through a magnetic field, the force exerted on it will be perpendicular to both the direction of its velocity and the magnetic field.
This perpendicular force is what gives rise to circular motion, as needed for centripetal force. It can maintain the particle's circular path by continually altering the direction of movement.
To calculate the magnetic force \( F_{m} \) acting on a charged particle, we use the equation:
  • \( F_{m} = qvB \sin(\theta) \)
Where:
  • \( q \) is the electric charge,
  • \( v \) is the velocity of the particle,
  • \( B \) is the magnetic field strength, and
  • \( \theta \) is the angle between the velocity direction and the magnetic field.
When the magnetic force serves as the centripetal force, the particle is confined to a circular path, which is why a magnetic field (\( B eq 0 \)) can allow circular motion without the presence of an electric field (\( E = 0 \)).
Electric Field
An electric field exerts a force on charged particles, directing them along the field lines. This force is described by Coulomb's law, which states that the force is proportional to the electric charge and the electric field strength.
For a particle in an electric field, this force acts linearly, not circularly. Thus, without a magnetic field, an electric field cannot provide the centripetal force necessary for a particle to follow a circular path.
The force \( F_{e} \) experienced by a charged particle in an electric field can be calculated using the formula:
  • \( F_{e} = qE \)
where:
  • \( q \) is the charge of the particle, and
  • \( E \) is the electric field's strength.
Any combination of electric and magnetic fields can be used to alter the trajectory of charged particles. However, independently, an isolated electric field cannot achieve or maintain circular motion unless it is paired with a magnetic force that provides the necessary centripetal force.

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Most popular questions from this chapter

A charged particle of specific charge (charge/mass) \(\alpha\) is released from origin at time \(t=0\) with velocity \(\vec{v}=v_{0}(i+\hat{j})\) in uniform magnetic field \(\bar{B}=B_{0} \hat{i}\). Coordinates of the particle at time \(t=\frac{\pi}{B_{0} \alpha}\) are \(\mathrm{a}\left(\frac{v_{0}}{2 B_{0} \alpha}, \frac{\sqrt{2} v_{0}}{\alpha B_{0}} ; \frac{-v_{0}}{B_{0} \alpha}\right)\) b \(\left(\frac{-v_{0}}{2 B_{0} \alpha}, 0 ; 0\right)\) c. \(\left(0, \frac{2 v_{0}}{B_{0} \alpha}, \frac{v_{0} \pi}{2 B_{0} \alpha}\right)\) \(d\left(\frac{v_{0} \pi}{B_{0} \alpha}, 0, \frac{-2 v_{0}}{B_{0} \alpha}\right)\)

An electron is moving along positive \(x\) -axis. A uniform electric field exists toward negative \(y\) -axis. What should be the direction of magnetic field of suitable magnitude so that net force on the electron is zero? a. Positive \(z\) -axis b Negative \(z\) -axis c. Positive \(y\) -axis d. Negative \(y\) -axis

The value of the electric field strength in vacuum if the energy density is same as that due to a magnetic field of induction \(1 \mathrm{~T}\) in vacuum is a. \(3 \times 10^{8} \mathrm{NC}^{-1}\) b \(1.5 \times 10^{8} \mathrm{NC}^{-1}\) c. \(2.0 \times 10^{8} \mathrm{NC}^{-1}\) d. \(1.0 \times 10^{8} \mathrm{NC}^{-1}\)

A long, straight, hollow conductor (tube) carrying a current has two sections \(A\) and \(C\) of unequal cross sections joined by a conical section \(B\). 1,2 and 3 are points on a line parallel to the axis of the conductor. The magnetic fields at 1,2 and 3 have magnitudes \(B_{1}, B_{2}\) and \(B_{3}\). Then, a. \(B_{1}=B_{2}=B_{3}\) b. \(B_{1}=B_{2} \neq B_{3}\) c. \(B_{1}

The circular coils \(A\) and \(B\) with their centers lying on the same axis have same number of turns and carry equal currents in the same sense. They are separated by a distance, have different diameters but subtend same angle at a point \(P\) lying on their common axis. The coil \(B\) lies exsctly midway between coil \(A\) and the point \(P\). The magnetic fiolds at point \(P\) due to coils \(A\) and \(B\) are \(B_{1}\) and \(B_{2}\), respectively. The ratio \(B_{1} / B_{2}\) is 8\. \(B_{1}>B_{2}\) b. \(B_{1}

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