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Chapter 9: Problem 19

A charged particle is whirled in a horizontal circle on a frictionless table by attaching it to a string fixed at one end. If a rnagnetic field is switched on in the vertical direction, the tension in the string a. will increase b. will decrease c. remains same d. may increase or decrease

Short Answer

Expert verified
b. will decrease

Step by step solution

01

Understand the Scenario

Consider a charged particle being rotated in a horizontal circle by a string. Initially, there is no magnetic field, so the centripetal force is provided by the tension in the string.
02

Analyze Magnetic Field Effect

When a magnetic field is applied in the vertical direction, it interacts with the moving charged particle. The magnetic force, which acts perpendicular to both the velocity and the magnetic field, adds a new horizontal component of force on the particle.
03

Determine Direction of Magnetic Force

Apply the right-hand rule to determine the force direction. If the magnetic force is oriented toward or away from the circle's center, it affects the required tension in the string accordingly.
04

Calculate Resulting Tension

Since the magnetic force adds an additional centripetal force component, the string needs less tension to keep the particle moving in a circle. Hence, the tension will decrease.

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Key Concepts

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

Charged Particle Motion
When charged particles move through a magnetic field, interactions between their velocity and the magnetic field create a force called the Lorentz force. This force is perpendicular to both the particle's velocity and the magnetic field direction. As a result, charged particles follow a curved path rather than a straight one.

In the given exercise, the charged particle is initially being swung in a circular motion on a table without a magnetic field. To maintain this circular motion, a force must counteract the particle's inertia that wants to keep it moving in a straight line. In absence of the magnetic field, this entire force is provided by the tension in the string.
  • Initial motion without the magnetic field relies solely on the tension in the string for centripetal force.
  • The introduction of the magnetic field affects this setup by providing an additional force element.
Understanding how charged particles interact with magnetic fields is crucial for explaining phenomena in electromagnetism and many technological applications, such as cyclotrons and mass spectrometers.
Centripetal Force
Centripetal force is the force that keeps an object moving in a circular path, directed towards the center of the circle around which the object is revolving. In the absence of other forces, the centripetal force must be completely supplied by the tension in the string for an object like a charged particle moving in a circular path. The mathematical expression for centripetal force is given by:\[ F_c = \frac{mv^2}{r} \]where:
  • \( F_c \) is the centripetal force
  • \( m \) is the mass of the particle
  • \( v \) is the velocity of the particle
  • \( r \) is the radius of the circle
Introduction of a magnetic field adds an additional force perpendicular to the velocity of the charged particle, thereby modifying the net centripetal force required to keep the particle moving in its circular path. If the magnetic component of the force assists the centripetal force, then the tension in the string would be reduced accordingly.
Tension in String
Tension in a string acts as the pulling force, maintaining the integrity of a system's motion when the object is connected to it. Tension ensures that the rotating particle remains in its path by providing the necessary centripetal force. With a magnetic field now acting on the charged particle, this scenario presents additional dynamics. The magnetic force interacts with the motion of the charged particle, introducing a component of force that adds to or reduces the needed centripetal force from the string.
  • If the magnetic force acts towards the center, it decreases the need for tension to provide centripetal force.
  • The tension required in the string is thus decreased, as part of the centripetal force is now supplied by the magnetic effect.
  • The balance between these forces determines new tension levels after the magnetic field is applied.
Understanding the tension in strings under such influences is crucial in mechanical systems where forces work in conjunction or opposition.

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Most popular questions from this chapter

A particle of charge \(-1.6 \times 10^{-18} \mathrm{C}\) moving with velocity 10 \(\mathrm{ms}^{-1}\) along the \(x\) -axis enters a region where a magnetic field of induction \(B\) is along the \(y\) -axis, and an electric ficld of magnitude \(10 \mathrm{Vm}^{-1}\) is along the negative \(z\) -axis. If the charged particle c枚ntinues moving along the \(x\) -axis, the magnitude of \(B\) is a. \(10^{-3} \mathrm{Wbm}^{-2}\) b \(10^{3} \mathrm{Wbm}^{-2}\) c. \(10^{2} \mathrm{Wbm}^{-2}\) d. \(10^{16} \mathrm{Wbm}^{-2}\)

A charged particle of mass \(10^{-3} \mathrm{~kg}\) and charge \(10^{-3}\) Centers a magnetic ficld of induction 1 tesia. If \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\), for what value of velocity will it pass straight through the field without deflection? a. \(10^{-3} \mathrm{~ms}^{-1}\) h \(10^{3} \mathrm{~ms}^{-1}\) c. \(10^{6} \mathrm{~ms}^{-1}\) d \(1 \mathrm{~ms}^{-1}\)

Two parallel wires carry currents of 20 and \(40 \mathrm{~A}\) in opposite directions. Another wire carrying a current antiparallel to 20 A is placed midway between the two wires. The magnetic force on it will be a toward \(20 \mathrm{~A} \quad \mathrm{~b}\) toward \(40 \mathrm{~A}\) c. zero d perpendicular to the plane of the currents

Four parallel conductors, carrying equal currents, pass. vertically through the four comers of a square WXYZ. In two conductors, the current is flowing into the page, and in the other two out of the page. In what directions must the currents flow to produce a resultant magnetic field in the direction shown at \(O\), the center of the square? \(9.278\) Into the page \(\quad\) Out of the page A \(W\) and \(Y\) and \(Z\) \(\mathbf{b} X\) and \(Z\) and \(Y\) c. \(W\) and \(Z\) \(X\) and \(Y\) d. \(W\) and \(X\) \(\boldsymbol{Y}\) and \(\mathrm{Z}\)

An electron of mass \(m\) is accelerated through a potential difference of \(V\) and then it enters a magnetic field of induction \(B\) normal to the lines. Then, the radius of the circular path is a. \(\sqrt{\frac{2 e V}{m}}\) b \(\sqrt{\frac{2 V m}{e B^{2}}}\) c. \(\sqrt{\frac{2 V m}{e B}}\) d. \(\sqrt{\frac{2 V m}{e^{2} B}}\)
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