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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

The magnitude of the magnetic force on a particle with charge \(-2 e\) moving with speed \(v=1.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) is \(3.0 \cdot 10^{-18} \mathrm{~N}\). What is the magnitude of the magnetic field component perpendicular to the direction of motion of the particle?

A straight wire with a constant current running through it is in Earth's magnetic field, at a location where the magnitude is \(0.43 \mathrm{G}\). What is the minimum current that must flow through the wire for a 10.0 -cm length of it to experience a force of \(1.0 \mathrm{~N} ?\)

A particle with mass \(m\) and charge \(q\) is moving within both an electric field and a magnetic field, \(\vec{E}\) and \(\vec{B}\). The particle has velocity \(\vec{v},\) momentum \(\vec{p}\), and kinetic energy, \(K\). Find general expressions for \(d \vec{p} / d t\) and \(d K / d t\), in terms of these seven quantities.

An electron is moving at \(v=6.00 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's magnetic field. If the field strength is \(0.500 \cdot 10^{-4} \mathrm{~T}\), what is the radius of the electron's circular path?

An electron (with charge \(-e\) and mass \(m_{\mathrm{e}}\) ) moving in the positive \(x\) -direction enters a velocity selector. The velocity selector consists of crossed electric and magnetic fields: \(\vec{E}\) is directed in the positive \(y\) -direction, and \(\vec{B}\) is directed in the positive \(z\) -direction. For a velocity \(v\) (in the positive \(x\) -direction \(),\) the net force on the electron is zero, and the electron moves straight through the velocity selector. With what velocity will a proton (with charge \(+e\) and mass \(m_{\mathrm{p}}=1836 \mathrm{~m}_{\mathrm{e}}\) ) move straight through the velocity selector? a) \(v\) b) \(-v\) c) \(v / 1836\) d) \(-v / 1836\)
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