/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 113 A proton moves in a helical path... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 113

A proton moves in a helical path at speed \(v=4.0\) $\times 10^{7} \mathrm{m} / \mathrm{s}$ high above the atmosphere, where Earth's magnetic field has magnitude \(B=1.0 \times 10^{-6} \mathrm{T}\). The proton's velocity makes an angle of \(25^{\circ}\) with the magnetic field. (a) Find the radius of the helix. [Hint: Use the perpendicular component of the velocity.] (b) Find the pitch of the helix-the distance between adjacent "coils." [Hint: Find the time for one revolution; then find how far the proton moves along a field line during that time interval.]

Short Answer

Expert verified
Question: Determine the radius of the helix and the pitch of the helix formed by the proton moving in the magnetic field. Solution: Step 1: Calculate the magnetic force on the proton: \(F = 1.6\times10^{-19}\,\text{C} \times (4.0\times10^7\,\text{m/s} \times \sin(25^\circ)) \times 1.0\times10^{-6}\,\text{T}\). Step 2: Calculate the centripetal force of the helix: \(F_c = \frac{1.67\times10^{-27}\,\text{kg} \times (4.0\times10^7\,\text{m/s} \times \sin(25^\circ))^2}{r}\). Step 3: Equate the magnetic force and centripetal force to find the radius of the helix: \(r = \frac{1.67\times10^{-27}\,\text{kg} \times (4.0\times10^7\,\text{m/s} \times \sin(25^\circ))}{1.6\times10^{-19}\,\text{C} \times 1.0\times10^{-6}\,\text{T}}\). Step 4: Calculate the time for one revolution: \(T = \frac{2\pi r}{4.0\times10^7\,\text{m/s} \times \sin(25^\circ)}\). Step 5: Calculate the pitch of the helix: \(P = (4.0\times10^7\,\text{m/s} \times \cos(25^\circ)) \times T\). Please perform the calculations based on the given formulas and report the numeric values for the radius and the pitch of the helix.

Step by step solution

01

(Step 1: Calculate the magnetic force on the proton)

(First, let's determine the magnetic force acting on the proton. The equation for magnetic force (F) on a charged particle is given by: \(F = q(v \times B)\), where q is the charge, v is the velocity, and B is the magnetic field. The charge of a proton is \(q = 1.6\times10^{-19}\,\text{C}\). The perpendicular component of the velocity can be calculated as \(v_{\perp} = v\sin\theta = 4.0\times10^7\,\text{m/s} \times \sin(25^\circ)\). Now, the magnetic force can be calculated as \(F = 1.6\times10^{-19}\,\text{C} \times (4.0\times10^7\,\text{m/s} \times \sin(25^\circ)) \times 1.0\times10^{-6}\,\text{T}\).)
02

(Step 2: Calculate the centripetal force of the helix)

(The magnetic force acting on the proton will cause it to move in a circular path. The centripetal force required for this motion can be expressed as \(F_c = \frac{mv_{\perp}^2}{r}\), where m is the mass of the proton, \(v_{\perp}\) is the perpendicular component of the velocity, and r is the radius of the helix. The mass of a proton is \(m = 1.67\times10^{-27}\,\text{kg}\). Substitute the values into the equation to find the centripetal force.)
03

(Step 3: Equate the magnetic force and centripetal force to find the radius of the helix)

(Since the magnetic force is the cause of the centripetal force acting on the proton, we can set the two forces equal to each other and solve for the radius of the helix. We get the equation: \(q(v_{\perp} \times B) = \frac{mv_{\perp}^2}{r}\). Solve this equation for r: \(r = \frac{mv_{\perp}}{qB}\). Substitute in the known values and calculate the radius.)
04

(Step 4: Calculate the time for one revolution)

(To find the time for one revolution, we can use the formula \(T = \frac{2\pi r}{v_{\perp}}\). Substitute the calculated radius and the perpendicular component of the velocity into the equation and solve for the time T.)
05

(Step 5: Calculate the pitch of the helix)

(The pitch of the helix is the distance between adjacent coils, and it can be calculated using the parallel component of velocity: \(v_{\parallel} = v\cos\theta = 4.0\times10^7\,\text{m/s} \times \cos(25^\circ)\). Then, we can calculate the distance the proton moves along the magnetic field line during one revolution using the formula: \(P = v_{\parallel} \times T\). Substitute the known values and calculate the pitch of the helix.)

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Most popular questions from this chapter

A magnet produces a \(0.30-\mathrm{T}\) field between its poles, directed to the east. A dust particle with charge \(q=-8.0 \times 10^{-18} \mathrm{C}\) is moving straight down at \(0.30 \mathrm{cm} / \mathrm{s}\) in this field. What is the magnitude and direction of the magnetic force on the dust particle?
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The torque on a loop of wire (a magnetic dipole) in a uniform magnetic field is \(\tau=N I A B\) sin \(\theta,\) where \(\theta\) is the angle between \(\overline{\mathbf{B}}\) and a line perpendicular to the loop of wire. Suppose an electric dipole, consisting of two charges \(\pm q\) a fixed distance \(d\) apart is in a uniform electric field \(\overrightarrow{\mathbf{E}}\). (a) Show that the net electric force on the dipole is zero. (b) Let \(\theta\) be the angle between \(\overrightarrow{\mathbf{E}}\) and a line running from the negative to the positive charge. Show that the torque on the electric dipole is \(\tau=q d E\) sin \(\theta\) for all angles $-180^{\circ} \leq \theta \leq 180^{\circ} .$ (Thus, for both electric and magnetic dipoles, the torque is the product of the dipole moment times the field strength times $\sin \theta .\( The quantity \)q d\( is the electric dipole moment; the quantity \)N I A$ is the magnetic dipole moment.)
Find the magnetic force exerted on an electron moving vertically upward at a speed of \(2.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) by a horizontal magnetic field of \(0.50 \mathrm{T}\) directed north.
The strip in the diagram is used as a Hall probe to measure magnetic fields. (a) What happens if the strip is not perpendicular to the field? Does the Hall probe still read the correct field strength? Explain. (b) What happens if the ficld is in the plane of the strip?
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