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Chapter 27: Problem 21

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?

Short Answer

Expert verified
Answer: The magnitude of the magnetic field component perpendicular to the direction of motion of the particle is approximately \(9.375 \times 10^{-5}\) T.

Step by step solution

01

Write down the given information

We are given that: - the particle has a charge \(q = -2e\), - the speed of the particle is \(v = 1.0 \cdot 10^5\) m/s, - the magnitude of the magnetic force is \(F_B = 3.0 \cdot 10^{-18}\) N.
02

Determine the angle between the velocity vector and the magnetic field vector

In this case, we are looking for the component of the magnetic field that is perpendicular to the direction of motion of the particle, so the angle \(\theta = 90^\circ\).
03

Use the formula for magnetic force to find the magnitude of the magnetic field

The formula for the magnetic force on a moving charged particle is: \(F_B = |q|(vB\sin{\theta})\) Plug in the known values: \(3.0 \cdot 10^{-18} \text{ N} = |-2e|(1.0 \cdot 10^5 \text{ m/s})(B\sin{90^\circ})\) We know that \(e = 1.6 \times 10^{-19} \text{ C}\) and \(\sin{90^\circ} = 1\). Therefore, the equation becomes: \(3.0 \cdot 10^{-18} \text{ N} = |-2(1.6 \times 10^{-19} \text{ C})|(1.0 \cdot 10^5 \text{ m/s})(B)(1)\)
04

Solve for the magnetic field \(B\)

Rearrange the formula to solve for \(B\): \(B = \frac{3.0 \times 10^{-18} \text{ N}}{-2(1.6 \times 10^{-19} \text{ C})(1.0 \times 10^5 \text{ m/s})}\) Now, compute the value of \(B\): \(B \approx 9.375 \times 10^{-5} \text{ T}\)
05

State the final answer

The magnitude of the magnetic field component perpendicular to the direction of motion of the particle is approximately \(9.375 \times 10^{-5}\) T.

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Most popular questions from this chapter

A rectangular coil with 20 windings carries a current of 2.00 mA flowing in the counterclockwise direction. It has two sides that are parallel to the \(y\) -axis and have length \(8.00 \mathrm{~cm}\) and two sides that are parallel to the \(x\) -axis and have length \(6.00 \mathrm{~cm} .\) A uniform magnetic field of \(50.0 \mu \mathrm{T}\) acts in the positive \(x\) -direction. What torque must be applied to the loop to hold it steady?

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 proton moving at speed \(v=1.00 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters a region in space where a magnetic field given by \(\vec{B}=\) \((-0.500 \mathrm{~T}) \hat{z}\) exists. The velocity vector of the proton is at an angle \(\theta=60.0^{\circ}\) with respect to the positive \(z\) -axis. a) Analyze the motion of the proton and describe its trajectory (in qualitative terms only). b) Calculate the radius, \(r\), of the trajectory projected onto a plane perpendicular to the magnetic field (in the \(x y\) -plane). c) Calculate the period, \(T,\) and frequency, \(f\), of the motion in that plane. d) Calculate the pitch of the motion (the distance traveled by the proton in the direction of the magnetic field in 1 period).

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?

A conducting rod of length \(L\) slides freely down an inclined plane, as shown in the figure. The plane is inclined at an angle \(\theta\) from the horizontal. A uniform magnetic field of strength \(B\) acts in the positive \(y\) -direction. Determine the magnitude and the direction of the current that would have to be passed through the rod to hold it in position on the inclined plane.
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