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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 simple galvanometer is made from a coil that consists of \(N\) loops of wire of area \(A .\) The coil is attached to a mass, \(M\), by a light rigid rod of length \(L\). With no current in the coil, the mass hangs straight down, and the coil lies in a horizontal plane. The coil is in a uniform magnetic field of magnitude \(B\) that is oriented horizontally. Calculate the angle from the vertical of the rigid rod as a function of the current, \(i\), in the coil.

In the Hall effect, a potential difference produced across a conductor of finite thickness in a magnetic field by a current flowing through the conductor is given by a) the product of the density of electrons, the charge of an electron, and the conductor's thickness divided by the product of the magnitudes of the current and the magnetic field. b) the reciprocal of the expression described in part (a). c) the product of the charge on an electron and the conductor's thickness divided by the product of the density of electrons and the magnitudes of the current and the magnetic field. d) the reciprocal of the expression described in (c). e) none of the above.

A small aluminum ball with a mass of \(5.00 \mathrm{~g}\) and a charge of \(15.0 \mathrm{C}\) is moving northward at \(3000 \mathrm{~m} / \mathrm{s}\). You want the ball to travel in a horizontal circle with a radius of \(2.00 \mathrm{~m}\), in a clockwise sense when viewed from above. Ignoring gravity, what is the magnitude and the direction of the magnetic field that must be applied to the aluminum ball to cause it to have this motion?

A 30 -turn square coil with a mass of \(0.250 \mathrm{~kg}\) and a side length of \(0.200 \mathrm{~m}\) is hinged along a horizontal side and carries a 5.00 -A current. It is placed in a magnetic field pointing vertically downward and having a magnitude of \(0.00500 \mathrm{~T}\). Determine the angle that the plane of the coil makes with the vertical when the coil is in equilibrium. Use \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

A particle with charge \(q\) is at rest when a magnetic field is suddenly turned on. The field points in the \(z\) -direction. What is the direction of the net force acting on the charged particle? a) in the \(x\) -direction b) in the \(y\) -direction c) The net force is zero. d) in the \(z\) -direction
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