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Chapter 27: Problem 23

A particle with a charge of \(20.0 \mu \mathrm{C}\) moves along the \(x\) -axis with a speed of \(50.0 \mathrm{~m} / \mathrm{s}\). It enters a magnetic field given by \(\vec{B}=0.300 \hat{y}+0.700 \hat{z},\) in teslas. Determine the magnitude and the direction of the magnetic force on the particle.

Short Answer

Expert verified
Answer: The magnitude of the magnetic force on the particle is \(300 \times 10^{-6} \, N\) and the direction is in the positive \(z\)-direction.

Step by step solution

01

Identify the given values

In this problem, we are given the following information: Charge of the particle, q: \(20.0\mu C\) Velocity of the particle, \(\vec{v}\): \(50.0 m/s\) in the \(x\)-direction Magnetic Field, \(\vec{B}\): \(0.300 \hat{y} + 0.700 \hat{z}\) T
02

Apply the Lorentz force equation

The Lorentz force, \(\vec{F}\), on a charged particle moving in a magnetic field is given by the equation: \(\vec{F} = q(\vec{v} \times \vec{B})\) In this case, \(\vec{v}\) and \(\vec{B}\) are only in the \(x\), \(y\), and \(z\) components, and we can write them as: \(\vec{v} = 50.0 \hat{x} \, \mathrm{(m/s)}\) \(\vec{B} = 0.300 \hat{y} + 0.700 \hat{z} \, \mathrm{(T)}\)
03

Compute the cross product \(\vec{v} \times \vec{B}\)

Calculate the cross product of the velocity vector \(\vec{v}\) and the magnetic field vector \(\vec{B}\): \((\vec{v} \times \vec{B}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 50.0 & 0 & 0 \\ 0 & 0.300 & 0.700 \end{vmatrix}\) Expanding the determinant, we get: \((\vec{v} \times \vec{B}) = (0\hat{i} - 0\hat{j} + 15\hat{k})\)
04

Calculate the magnetic force \(\vec{F}\)

Now, plug the cross product and charge q into the Lorentz force equation to find the magnetic force: \(\vec{F} = (20.0 \times 10^{-6})(0\hat{i} - 0\hat{j} + 15\hat{k})\) \(\vec{F} = 300 \times 10^{-6} \hat{k} \, N\)
05

Find the magnitude and direction of the magnetic force

The magnitude of the magnetic force is given by the expression: \(|\vec{F}| = 300 \times 10^{-6} \,N\) Now, we can determine the direction of the magnetic force. Since the force only has a component in the \(z\)-direction, the force acts along the \(z\)-axis. So, the magnitude of the magnetic force on the particle is \(300 \times 10^{-6} \, N\) in the positive \(z\)-direction.

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Most popular questions from this chapter

A current-carrying wire is positioned within a large, uniform magnetic field, \(\vec{B}\). However, the wire experiences no force. Explain how this might be possible.

A magnetic field is oriented in a certain direction in a horizontal plane. An electron moves in a certain direction in the horizontal plane. For this situation, there a) is one possible direction for the magnetic force on the electron. b) are two possible directions for the magnetic force on the electron. c) are infinite possible directions for the magnetic force on the electron.

A rail gun accelerates a projectile from rest by using the magnetic force on a current-carrying wire. The wire has radius \(r=5.1 \cdot 10^{-4} \mathrm{~m}\) and is made of copper having a density of \(\rho=8960 \mathrm{~kg} / \mathrm{m}^{3}\). The gun consists of rails of length \(L=1.0 \mathrm{~m}\) in a constant magnetic field of magnitude \(B=2.0 \mathrm{~T}\) oriented perpendicular to the plane defined by the rails. The wire forms an electrical connection across the rails at one end of the rails. When triggered, a current of \(1.00 \cdot 10^{4}\) A flows through the wire, which accelerates the wire along the rails. Calculate the final speed of the wire as it leaves the rails. (Neglect friction.)

A circular coil with a radius of \(10.0 \mathrm{~cm}\) has 100 turns of wire and carries a current, \(i=100 . \mathrm{mA} .\) It is free to rotate in a region with a constant horizontal magnetic field given by \(\vec{B}=(0.0100 \mathrm{~T}) \hat{x}\). If the unit normal vector to the plane of the coil makes an angle of \(30.0^{\circ}\) with the horizontal, what is the magnitude of the net magnetic torque acting on the coil? 27.61 At \(t=0\) an electron crosses the positive \(y\) -axis (so \(x=0\) ) at \(60.0 \mathrm{~cm}\) from the origin with velocity \(2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. It is in a uniform magnetic field. a) Find the magnitude and the direction of the magnetic field that will cause the electron to cross the \(x\) -axis at \(x=60.0 \mathrm{~cm}\). b) What work is done on the electron during this motion? c) How long will the trip take from \(y\) -axis to \(x\) -axis?

A proton moving with a speed of \(4.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(y\) -direction enters a uniform magnetic field of \(0.40 \mathrm{~T}\) pointing in the positive \(x\) -direction. Calculate the magnitude of the force on the proton.
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