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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

Initially at rest, a small copper sphere with a mass of \(3.00 \cdot 10^{-6} \mathrm{~kg}\) and a charge of \(5.00 \cdot 10^{-4} \mathrm{C}\) is accelerated through a \(7000 .-\mathrm{V}\) potential difference before entering a magnetic field of magnitude \(4.00 \mathrm{~T}\), directed perpendicular to its velocity. What is the radius of curvature of the sphere's motion in the magnetic field?

Show that the magnetic dipole moment of an electron orbiting in a hydrogen atom is proportional to its angular momentum, \(L: \mu=-e L / 2 m,\) where \(-e\) is the charge of the electron and \(m\) is its mass.

It would be mathematically possible, for a region with zero current density, to define a scalar magnetic potential analogous to the electrostatic potential: \(V_{B}(\vec{r})=-\int_{\vec{r}_{0}}^{\vec{r}} \vec{B} \cdot d \vec{s},\) or \(\vec{B}(\vec{r})=-\nabla V_{B}(\vec{r}) .\) However, this has not been done. Explain why not.

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.

A coil consists of 120 circular loops of wire of radius \(4.8 \mathrm{~cm} .\) A current of 0.49 A runs through the coil, which is oriented vertically and is free to rotate about a vertical axis (parallel to the \(z\) -axis). It experiences a uniform horizontal magnetic field in the positive \(x\) -direction. When the coil is oriented parallel to the \(x\) -axis, a force of \(1.2 \mathrm{~N}\) applied to the edge of the coil in the positive \(y\) -direction can keep it from rotating. Calculate the strength of the magnetic field.
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