/*! 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 163 A charged particle of mass \(m\)... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 163

A charged particle of mass \(m\) and charge \(q\) travels on a circular path of radius \(r\) that is perpendicular to complete one revolution is (A) \(\frac{2 \pi q^{2} B}{m}\) (B) \(\frac{2 \pi m q}{B}\) (C) \(\frac{2 \pi m}{q B}\) (D) \(\frac{2 \pi q B}{m}\)

Short Answer

Expert verified
The time required for a charged particle of mass \(m\) and charge \(q\) to travel on a circular path of radius \(r\) that is perpendicular to complete one revolution is (C) \(\frac{2 \pi m}{q B}\).

Step by step solution

01

Understanding the key theories

A charged particle moving in a magnetic field experiences a magnetic force given by \(F = q(vB\sin\theta)\), where \(F\) is the magnetic force, \(q\) is the charge of the particle, \(v\) is the particle's velocity, \(B\) is the magnetic field's strength, and \(\theta\) is the angle between the particle's velocity and the magnetic field. In this case, the velocity is perpendicular to the magnetic field (\(\theta = 90^{\circ}\)), so we have \(\sin\theta = 1\). The magnetic force experienced by a charged particle moving perpendicular to a magnetic field acts as the centripetal force, which is given by \(F_c = \frac{mv^2}{r}\), where m is the mass of the particle, v is its velocity, and r is the radius of the circular path.
02

Setting up the equation

We can now equate the centripetal force with the magnetic force: \[\frac{mv^2}{r} = qvB\]
03

Finding the velocity

First, let's find the velocity of the charged particle: \[v = \frac{qrB}{m}\] Since the particle is traveling in a circular path, we can also write the distance it covers as \(D = 2\pi r\).
04

Finding the time for one revolution

Now let's find the time required to complete one revolution. The time (\(t\)) can be found using the formula \(D = vt\): \[2\pi r = \left(\frac{qrB}{m}\right)t\] Solving for \(t\), we get: \[t = \frac{2\pi m}{qB}\] Comparing this result with the given options, we can see that the correct answer is: (C) \(\frac{2 \pi m}{q B}\)

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Most popular questions from this chapter

Two short bar magnets of magnetic moments \(M\) each are arranged at the opposite corners of a square of side \(d\) such that their centres coincide with the corners and their axes are parallel. If the like poles are in the same direction, the magnetic induction at any of the other corners of the square is (A) \(\frac{\mu_{0}}{4 \pi} \frac{M}{d^{3}}\) (B) \(\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}}\) (C) \(\frac{\mu_{0}}{4 \pi} \frac{M}{2 d^{3}}\) (D) \(\frac{\mu_{0}}{4 \pi} \frac{M^{2}}{2 d^{3}}\)

Two concentric coils each of radius equal to \(2 \pi \mathrm{cm}\) are placed at right angles to each other, \(3 \mathrm{~A}\) and \(4 \mathrm{~A}\) are the currents flowing in coils, respectively. The magnetic induction in \(\mathrm{Wb} / \mathrm{m}^{2}\) at the centre of the coils will be \(\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{~Wb} / \mathrm{Am}\right)\)(A) \(12 \times 10^{-5}\) (B) \(10^{-5}\) (C) \(5 \times 10^{-5}\) (D) \(7 \times 10^{-5}\)

A magnetic needle is kept in a non-uniform magnetic field. It experiences (A) a torque but not a force. (B) neither a force nor a torque. (C) a force and a torque. (D) a force but not a torque.

A flat coil \(A B C D\), of \(n\) turns, area \(A\), and resistance \(R\) is placed in a uniform magnetic field of magnitude \(B_{0}\). The plane of the coil is initially perpendicular to magnitude field \(B_{0}\). If the coil is rotated by an angle \(\theta\) about the axis \(X Y\) (passing through centre and parallel to \(A D\) ), charge of amount \(Q\) flows through it. (A) If \(\theta=90^{\circ}, Q=\frac{B A n}{R}\) (B) If \(\theta=180^{\circ}, Q=\frac{B A n}{R}\)A flat coil \(A B C D\), of \(n\) turns, area \(A\), and resistance \(R\) is placed in a uniform magnetic field of magnitude \(B_{0}\). The plane of the coil is initially perpendicular to magnitude field \(B_{0}\). If the coil is rotated by an angle \(\theta\) about the axis \(X Y\) (passing through centre and parallel to \(A D\) ), charge of amount \(Q\) flows through it. (A) If \(\theta=90^{\circ}, Q=\frac{B A n}{R}\) (B) If \(\theta=180^{\circ}, Q=\frac{B A n}{R}\)

When a long wire carrying a steady current is bent into a circular coil of one turn, the magnetic induction at its centre is \(B\). When the same wire carrying the same current is bent to form a circular coil of \(n\) turns of a smaller radius, the magnetic induction at the centre will be (A) \(B / n\) (B) \(n B\) (C) \(B / n^{2}\) (D) \(n^{2} B\)
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