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Chapter 27: Problem 28

The Earth is showered with particles from space known as muons. They have a charge identical to that of an electron but are many times heavier \(\left(m=1.88 \cdot 10^{-28} \mathrm{~kg}\right)\) Suppose a strong magnetic field is established in a lab \((B=0.50 \mathrm{~T})\) and a muon enters this field with a velocity of \(3.0 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) at a right angle to the field. What will be the radius of the resulting orbit of the muon?

Short Answer

Expert verified
In this exercise, a muon particle with mass \(1.88 \times 10^{-28}\text{ kg}\) and velocity \(3.0 \times 10^6\text{ m/s}\) enters a magnetic field of strength \(0.50 \text{ T}\) at a right angle. Calculate the radius of the resulting orbit. The muon's orbit radius is approximately \(9.47\times 10^{-3}\text{ m}\) or \(9.47\text{ mm}\).

Step by step solution

01

Calculate the magnetic force on the muon

Since the muon enters the magnetic field at a right angle, the force acting on it due to the magnetic field can be calculated using the formula: \(F = qvB\sin(\theta)\) As the muon has an identical charge as that of an electron, \(q=-1.60 \times 10^{-19} \text{C}\) and given \(\theta=90^{\circ}\) and \(\sin(90^{\circ})=1\), so: \(F = (-1.60 \times 10^{-19})(3.0 \times 10^6)(0.50)\) \(F = 2.40 \times 10^{-13} \text{N}\)
02

Equate the magnetic force to the centripetal force

The magnetic force acting on the muon is equal to the centripetal force required to keep the muon in an orbit: \(F = \frac{mv^2}{r}\) Where \(m=1.88 \times 10^{-28}\text{ kg}\), \(v=3.0 \times 10^6\text{ m/s}\) and \(r\) denotes the radius. We can substitute the values and equate the two forces: \(2.40 \times 10^{-13} = \frac{(1.88 \times 10^{-28})(3.0 \times 10^6)^2}{r}\)
03

Calculate the radius of the muon's orbit

Now, we will solve for the radius \(r\): \(r=\frac{(1.88 \times 10^{-28})(3.0 \times 10^6)^2}{2.40 \times 10^{-13}}\) \(r = 9.47 \times 10^{-3} \text{m}\) So the radius of the resulting orbit of the muon is \(9.47\times 10^{-3}\text{ m}\) or \(9.47\text{ mm}\).

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Most popular questions from this chapter

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 a charge of \(+10.0 \mu \mathrm{C}\) is moving at \(300 \cdot \mathrm{m} / \mathrm{s}\) in the positive \(z\) -direction. a) Find the minimum magnetic field required to keep it moving in a straight line at constant speed if there is a uniform electric field of magnitude \(100 . \mathrm{V} / \mathrm{m}\) pointing in the positive \(y\) -direction. b) Find the minimum magnetic field required to keep the particle moving in a straight line at constant speed if there is a uniform electric field of magnitude \(100 . \mathrm{V} / \mathrm{m}\) pointing in the positive \(z\) -direction.

An electron moves in a circular trajectory with radius \(r_{\mathrm{i}}\) in a constant magnetic field. What is the final radius of the trajectory when the magnetic field is doubled? a) \(\frac{r_{i}}{4}\) b) \(\frac{r_{i}}{2}\) c) \(r_{i}\) d) \(2 r_{i}\) e) \(4 r_{\mathrm{i}}\)

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.

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.
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