/*! 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 34 A uniform magnetic field, \(B=B_... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A uniform magnetic field, \(B=B_{0} \hat{\mathbf{j}}\) exists in space. A particle of mass \(m\) and charge, \(q\) is projected towards \(x\)-axis with speed, \(v\) from a point \((a, 0,0)\). The maximum value of \(v\) for which the particle does not hit the \(y z\)-plane is (a) \(\frac{B q a}{m}\) (b) \(\frac{\mathrm{Bq} a}{2 \mathrm{~m}}\) (c) \(\frac{B q}{a m}\) (d) \(\frac{B q}{2 a m}\)

Short Answer

Expert verified
The maximum value of \(v\) for which the particle does not hit the yz-plane is (b) \(\frac{B q a}{2 m}\).

Step by step solution

01

Understanding the Motion

The particle is moving in a magnetic field that is perpendicular to its velocity. This results in a circular motion. The initial velocity is along the x-axis, and the magnetic field is along the y-axis.
02

Determine the Radius of Circular Motion

The centripetal force required for circular motion is provided by the magnetic force. The centripetal force is given by \(F_c = \frac{mv^2}{r}\), and the magnetic force is \(F_m = qvB\). Equating them, \(\frac{mv^2}{r} = qvB\). Thus, the radius of the path is \(r = \frac{mv}{qB}\).
03

Condition for Not Hitting yz-plane

The particle will hit the yz-plane if its path exceeds half a circle or \(\pi r\). Therefore, for the particle to not hit the yz-plane after it is projected, the distance \(a\) must be greater than or equal to \(\frac{rc}{2}\). Substitute the radius expression, \(a = \frac{mv}{qB / 2}\) gives \(v = \frac{qBa}{2m}\).
04

Compare with Given Options

From the derived velocity, \(v = \frac{qBa}{2m}\), we see that this 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.

Circular Motion
When we talk about circular motion, we're referring to the movement of a particle along a circular path. Think of a particle tethered to an invisible string, moving in a circle around a central point. In this scenario, the particle's direction is continuously changing, even if its speed remains constant.
This creates a special type of velocity known as "tangential velocity", which is always directed along the tangent to the circle at the particle's position. Since the direction is constantly changing, even though the speed is constant, the motion is considered accelerated.
  • This acceleration is due to a force that acts towards the center of the circle, keeping the particle on its curved path.
  • In our situation with the uniform magnetic field, it's this central-seeking force that's crucial for maintaining the circular motion of the charged particle.
Magnetic Force
Magnetic force is a crucial player in guiding the path of charged particles in a magnetic field. This force acts perpendicular to the velocity of the particle and changes only the direction of the particle, not its speed.
  • In terms of physics, when a charged particle moves in a magnetic field, it experiences a magnetic force perpendicular to both its velocity and the magnetic field.
  • The magnitude of this force is given by the formula: \( F_m = qvB \).
Where
  • \( q \) is the charge of the particle,
  • \( v \) is the velocity, and
  • \( B \) is the magnetic field strength.
This force is constant across the particle's motion, maintaining its speed but directing it in a circular path. The force acts as the cause of circular motion, providing the necessary centripetal force to keep the particle in its orbit.
Centripetal Force
Centripetal Force is the glue that keeps a particle moving in a circle. This inward force ensures that an object follows a circular trajectory rather than continuing in a straight line. For any object traveling in circular motion, like our particle in the magnetic field, centripetal force is essential.
In our magnetic field exercise, the centripetal force is supplied by the magnetic force. Here's how it works:
  • The centripetal force needed for the circular motion is given by the formula: \( F_c = \frac{mv^2}{r} \).
  • Equating the centripetal force and the magnetic force, we get: \( \frac{mv^2}{r} = qvB \).
  • Solving for the radius, \( r = \frac{mv}{qB} \), we find that the radius of the circle depends on the mass of the particle, its velocity, the charge, and the magnetic field strength.
This force ensures the particle doesn't spiral off into random space and stays neatly within its circular path. Understanding this relationship helps us determine conditions for motion, such as ensuring a particle does not hit undesired planes like the yz-plane in the problem at hand.

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

An infinitely long wire carrying current \(i\) is along \(Y\)-axis such that its one end is at point \((0, b)\) while the wire extends upto \(\infty .\) The magnitude of magnetic field strength at point \(P(a, 0)\) is (a) \(\frac{\mu_{0} i}{4 \pi a}\left(1+\frac{b}{\sqrt{a^{2}+b^{2}}}\right)\) (b) \(\frac{\mu_{0} i}{4 \pi a}\left(1-\frac{b}{\sqrt{a^{2}+b^{2}}}\right)\) (c) \(\frac{\mu_{0} i}{4 \pi a}\left(1-\frac{a}{\sqrt{a^{2}+b^{2}}}\right)\) (d) \(\frac{\mu_{0} i}{4 \pi a}\left(1+\frac{a}{\sqrt{a^{2}+b^{2}}}\right)\)

The magnetic field normal to the plane of a wire of \(n\) turns and radius \(r\) which carries a current \(i\) is measured on the axis of the coil at a small distance \(h\) from the centre of the coil. This is smaller than the magnetic field at the centre by the fraction (a) \((2 / 3) r^{2} / h^{2}\) (b) \((3 / 2) r^{2} / h^{2}\) (c) \((2 / 3) h^{2} / r^{2}\) (d) \((3 / 2) h^{2} / r^{2}\)

Two insulating plates are both uniformly charged in such a way that the potential difference between them is \(V_{2}-V_{1}=20 \mathrm{~V}\) (i.e, plate 2 is at a higher potential). The plates are separated by \(d=0.1 \mathrm{~m}\) and can be treated as infinitely large. An electron is released from rest on the inner surface of plate \(1 .\) What is its speed when it hits plate 2 ? [AIEEE 2006] (a) \(32 \times 10^{-19} \mathrm{~ms}^{-1}\) (b) \(2.65 \times 10^{6} \mathrm{~ms}^{-1}\) (c) \(7.02 \times 10^{12} \mathrm{~ms}^{-1}\) (d) \(1.87 \times 10^{6} \mathrm{~ms}^{-1}\)

The resistance of the shunt required to allow \(2 \%\) of the main current through the galvanometer of \mathrm{\\{} r e s i s t a n c e ~ \(49 \Omega\) is \(\quad\) [Kerala CET 2008] (a) \(1 \Omega\) (b) \(2 \Omega\) (c) \(0.2 \Omega\) (d) \(0.1 \Omega\) (e) 0010

A long straight, solid metal wire of radius \(2 \mathrm{~mm}\) carries a current uniformly distributed over its circular cross-section. The magnetic field induction at a distance \(2 \mathrm{~mm}\) from its axis is \(B\). Then, the magnetic field induction at distance \(1 \mathrm{~mm}\) from axis will be (a) \(B\) (b) \(B_{1} 2\) (c) \(2 \underline{B}\) (d) \(B\)

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.