/*! 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 25 A particle of charge \(-16 \time... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 25

A particle of charge \(-16 \times 10^{-18} \mathrm{C}\) moving with velocity \(10 \mathrm{~m} / \mathrm{s}\) along the \(x\)-axis enters a region where a magnetic field of induction \(B\) is along the \(y\)-axis and an electric field of magnitude \(10^{4} \mathrm{~V} / \mathrm{m}\) is along the negative \(z\)-axis. If the charged particle continues moving along the \(x\)-axis, the magnitude of \(B\) is (A) \(10^{3} \mathrm{~Wb} / \mathrm{m}^{2}\) (B) \(10^{5} \mathrm{~Wb} / \mathrm{m}^{2}\) (C) \(10^{16} \mathrm{~Wb} / \mathrm{m}^{2}\) (D) \(10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}\)

Short Answer

Expert verified
The magnitude of the magnetic field induction B is \(10^{-3} \, \mathrm{ Wb/m^2}\).

Step by step solution

01

Identify the given information

We are given the following information: 1. Charge of the particle, \(q = -16 \times 10^{-18} \, \mathrm{C}\) 2. Velocity of the particle, \(v = 10 \, \mathrm{m/s}\) along the x-axis 3. Electric field magnitude, \(E = 10^4 \, \mathrm{V/m}\) along the negative z-axis 4. Magnetic field direction is along the y-axis, while the magnitude B is unknown
02

Calculate the electric force acting on the particle

Since we have the electric field magnitude and the charge of the particle, we can find the electric force acting on the particle. The formula for the electric force is given by: \[F_{E} = qE\] Now, calculate the electric force acting on the particle: \(F_{E} = (-16 \times 10^{-18} \, \mathrm{C})(10^{4} \,\mathrm{V/m}\))
03

Set up the Lorentz force equation

We know that the Lorentz force acting on the charged particle is the sum of the electric and magnetic forces. The particle moves in a straight line, so the net force (F_net) acting on the particle should be zero. The formula for the Lorentz force is: \[F_{net} = F_{E} + F_{B} = 0\] It is important to note that since the magnetic field B is along the y-axis and the electric field is along the negative z-axis, the forces are acting perpendicular to each other.
04

Calculate the magnetic force acting on the particle

Since we know that the net force is zero, we can calculate the magnetic force acting on the particle as follows: \[F_B = -F_E\] Using the formula for the magnetic force: \[F_B = qvB\] Now, substitute the calculated value of the electric force: \[qvB = -F_E\]
05

Solve for the magnitude of magnetic field B

We have the equation: \[qvB = -F_E\] Now, we need to solve for the magnitude of the magnetic field (B). First, divide both sides of the equation by the charge (q) and the velocity (v): \[B = -\frac{F_E}{qv}\] Now, substitute the calculated values for the electric force (F_E), charge (q), and velocity (v): \[B = -\frac{(-16 \times 10^{-18} \, \mathrm{C})(10^{4} \, \mathrm{V/m})}{(-16 \times 10^{-18} \, \mathrm{C})(10 \, \mathrm{m/s})}\] Calculating the value of B, we get: \[B = 10^{-3} \, \mathrm{ Wb/m^2}\] The magnitude of the magnetic field induction B is \(10^{-3} \, \mathrm{ Wb/m^2}\), which corresponds to the option (D).

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Most popular questions from this chapter

Three identical bar magnets, each of magnetic moment \(M\), are placed in the form of an equilateral triangle with north pole of one touching the south pole of the other as shown. The net magnetic moment of the system is (A) Zero (B) \(3 M\) (C) \(\frac{3 M}{2}\) (D) \(M \sqrt{3}\)

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current \(i\). The magnetic field at its centre is \(6.28 \times 10^{-2}\) \(\mathrm{Wb} / \mathrm{m}^{2}\) Another long solenoid has 100 turns per \(\mathrm{cm}\) and it carries a current \(\frac{i}{3} .\) The value of the magnetic field at its centre is (A) \(1.05 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^{2}\) (B) \(1.05 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\) (C) \(1.05 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}\) (D) \(1.05 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\)

Two circular coils of radii \(5 \mathrm{~cm}\) and \(10 \mathrm{~cm}\) carry equal currents of \(2 \mathrm{~A}\). The coils have 50 and 100 turns, respectively, and are placed in such a way that their planes and their centres coincide. Magnitude of magnetic field at the common centre of coils is, (A) \(8 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in same direction. (B) \(4 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in opposite direction. (C) zero, if currents in the coils are in opposite direction. (D) \(8 \pi \times 10^{-4} \mathrm{~T}\) if currents in the coil are in opposite direction.

A current \(i\) ampere flows in a circular arc of wire whose radius is \(R\), which subtend an angle \(3 \pi / 2\) radian at its centre. The magnetic induction \(B\) at the centre is (A) \(\frac{\mu_{0} i}{R}\) (B) \(\frac{\mu_{0} i}{2 R}\) (C) \(\frac{2 \mu_{0} i}{R}\) (D) \(\frac{3 \mu_{0} i}{8 R}\)

Relative permittivity and permeability of a material \(\varepsilon_{r}\) and \(\mu_{r}\) respectively. Which of the following value of these quantities are allowed for a diamagnetic material? (A) \(\varepsilon_{r}=0.5, \mu_{r}=1.5\) (B) \(\varepsilon_{r}=1.5, \mu_{r}=0.5\) (C) \(\varepsilon_{r}=0.5, \mu_{r}=0.5\) (D) \(\varepsilon_{r}=1.5, \mu_{r}=1.5\)
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