/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 A pulsar is a neutron star havin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A pulsar is a neutron star having magnetic field at \(10^{12} \mathrm{G}\) at its surface. The maximum magnetic force experienced by an electron moving with velocity \(0.9 c\) is (a) \(43.2 \mathrm{~N}\) (b) \(4.32 \times 10^{-3} \mathrm{~N}\) (c) \(4.32 \times 10^{3} \mathrm{~N}\) (d) zero

Short Answer

Expert verified
(b) \(4.32 \times 10^{-3} \, \mathrm{N}\)

Step by step solution

01

Determine the formula for magnetic force

The formula for the magnetic force experienced by a charged particle moving in a magnetic field is given by the Lorentz force equation: \( F = qvB \sin(\theta) \) 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, and \( \theta \) is the angle between the velocity and the magnetic field. For maximum force, \( \sin(\theta) = 1 \).
02

Identify known values

From the problem statement, we identify the known values: - The magnetic field at the surface of the pulsar, \( B = 10^{12} \, \text{G} \), which needs to be converted to Tesla. Since \( 1 \, \text{G} = 10^{-4} \, \text{T} \), this becomes \( B = 10^{8} \, \text{T} \).- The charge of an electron is \( q = -1.6 \times 10^{-19} \text{C} \).- The velocity of the electron \( v = 0.9c \), where \( c = 3 \times 10^{8} \text{m/s} \). Hence, \( v = 0.9 \times 3 \times 10^{8} = 2.7 \times 10^{8} \text{m/s} \).
03

Calculate the maximum force

Substitute the known values into the equation for maximum force: \[ F = qvB \sin(90^\circ) = qvB \]Substitute \( q = 1.6 \times 10^{-19} \text{C} \), \( v = 2.7 \times 10^{8} \text{m/s} \), and \( B = 10^{8} \text{T} \) into the equation:\[ F = (1.6 \times 10^{-19}) \times (2.7 \times 10^{8}) \times (10^{8}) \]\[ F = 4.32 \times 10^{-3} \text{N} \]
04

Compare calculated force with given options

The calculated force is \( 4.32 \times 10^{-3} \text{N} \), which matches option (b).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Neutron Star
Neutron stars are the remnants of massive stars that have undergone a supernova explosion. These stars are incredibly dense and compact, often with masses greater than that of the sun. However, their radius is only about 10 kilometers, which is comparable to the size of a city. The density of a neutron star is so immense that a sugar-cube-sized amount of its material would weigh about the same as Mount Everest.

Neutron stars are composed primarily of neutrons and have strong gravitational fields. This gravity is so powerful that it significantly warps space-time around the star. Additionally, neutron stars possess intense magnetic fields, much stronger than those typically found on Earth. These magnetic fields can reach strengths of up to a trillion times those of Earth's magnetic field.

Pulsars are a specific kind of neutron star that emit beams of electromagnetic radiation from their magnetic poles. As the star rotates, these beams sweep through space, often detectable as regular pulses of radiation. This specific exercise involves a pulsar, highlighting the extreme magnetic fields characteristic of neutron stars.
Lorentz Force
The Lorentz force is the force experienced by a charged particle moving in a magnetic field. It is named after the Dutch physicist Hendrik Lorentz and plays a critical role in magnetism and electromagnetism. The formula for calculating this force is given by the equation: \[ F = qvB \sin(\theta) \]where:
  • \( F \) is the magnetic force.
  • \( q \) represents the charge of the particle.
  • \( v \) indicates the velocity of the particle.
  • \( B \) symbolizes the magnetic field strength.
  • \( \theta \) is the angle between the velocity vector and the magnetic field vector.

When the velocity of the particle is perpendicular to the magnetic field, \( \theta = 90^\circ \) and \( \sin(\theta) = 1 \), resulting in the maximum force experienced by the particle. Hence, for the exercise, we consider the scenario wherein the electron moves at this ideal angle to maximize the force.

The Lorentz force causes charged particles to follow a curved path when moving through a magnetic field. This principle is extensively used in devices like cyclotrons and in phenomena such as the auroras, where charged particles from the sun interact with Earth's magnetic field.
Magnetic Field Strength
Magnetic field strength is a critical parameter in the study of magnetic forces. It is denoted by \( B \) and measured in Tesla (T). The strength of a magnetic field determines how much force it can exert on a charged particle that is moving through it.

In this exercise, the neutron star's magnetic field strength is given as \( 10^{12} \mathrm{G} \). To convert Gauss (G) to Tesla, we use the conversion factor where \( 1 \, \text{G} = 10^{-4} \, \text{T} \). Therefore, the magnetic field strength at the neutron star's surface becomes \( 10^8 \mathrm{T} \). Such a powerful magnetic field can significantly affect the trajectory of charged particles moving near the star.

Understanding magnetic field strength is crucial in both astrophysics and everyday applications on Earth. This concept helps engineers design technologies such as magnetic resonance imaging (MRI) machines and magnetic levitation systems, which rely on controlled magnetic forces.
Velocity of Charged Particle
The velocity of a charged particle is key in determining the magnetic force acting upon it. Velocity not only affects the magnitude of the force but also its direction, as per the Lorentz force equation.

For calculation purposes, we understand velocity as the measure of how fast a particle moves in a given direction. In our scenario, we deal with an electron moving at a velocity equal to \( 0.9c \), where \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \text{m/s} \). Thus, the velocity \( v \) of the electron is calculated to be \( 2.7 \times 10^8 \text{m/s} \).

The velocity of a particle can drastically alter not just the magnetic force but also its path through the magnetic field. This principle is evident in particle accelerators, where velocities approaching the speed of light enhance the energy and behavior of particles, thereby helping scientists study fundamental interactions in physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Assertion Out of galvanometer, ammeter and voltmeter, resistance of ammeter is lowest and resistance of voltmeter is highest. Reason An ammeter is connected in series and a voltmeter is connected in parallel, in a circuit.

A rectangular loop carrying current is placed near a long straight fixed wire carrying strong current such that long sides are parallel to wire. If the current in the nearer long side of loop is parallel to current in the wire. Then the loop (a) experiences no force (b) experiences a force towards wire (c) experiences a force away from wire (d) experiences a torque but no force

Biot-Savart law indicates that the moving electrons (velocity \(\mathbf{v}\) ) produce a magnetic field \(\mathbf{B}\) such that [NCERT Exemplar] (a) \(\mathrm{B} \perp \mathrm{v}\) (b) \(\mathrm{B} \| \mathrm{v}\) (c) it obeys inverse cube law (d) it is along the line joining the electron and point of observation

A circular coil of 5 turns and of \(10 \mathrm{~cm}\) mean diameter is connected to a voltage source. If the resistance of the coil is \(10 \mathrm{~W}\), the voltage of the source so as to nullify the horizontal component of earth's magnetic field of \(30 \mathrm{~A}\) turn \(\mathrm{m}^{-1}\) at the centre of the coil should be \(\quad\) [Kerala CET 2007] (a) \(6 \mathrm{~V}\), plane of the coil normal to the magnetic meridian (b) \(2 \mathrm{~V}\), plane of coil normal to the magnetic meridian (c) \(2 \mathrm{~V}\), plane of the coil along the magnetic meridian (d) \(4 \mathrm{~V}\), plane of the coil normal to magnetic meridian

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current \(i .\) The magnetic field at its centre is \(6.28 \times 10^{-2} \mathrm{Wbm}^{-2} .\) Another along solenoid has 100 turns per \(\mathrm{cm}\) and it carries a current \(i / 3 .\) The value of the magnetic field at its centre [AIEEE 2006] (a) \(1.05 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}\) (b) \(1.05 \times 10^{-2} \mathrm{~Wb} \mathrm{~m}^{-2}\) (c) \(1.05 \times 10^{-5} \mathrm{~Wb} \mathrm{~m}^{-2}\) (d) \(1.05 \times 10^{-3} \mathrm{~Wb} \mathrm{~m}^{-2}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.