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Chapter 27: Problem 30

An electron with energy equal to \(4.00 \cdot 10^{2} \mathrm{eV}\) and an electron with energy equal to \(2.00 \cdot 10^{2} \mathrm{eV}\) are trapped in a uniform magnetic field and move in circular paths in a plane perpendicular to the magnetic field. What is the ratio of the radii of their orbits?

Short Answer

Expert verified
Answer: The ratio of the radii of the orbits of the two electrons is √2.

Step by step solution

01

Formula for the radius of the circular orbit

The formula for the radius (r) of the circular orbit an electron with charge (q) and velocity (v) moving in a plane perpendicular to a magnetic field (B) is given by: r = mv/(qB), where m is the mass of the electron.
02

Formula for the energy of the electron

The energy (E) of an electron is given by the formula: E = (1/2)mv^2
03

Express the velocity in terms of energy

Solving the energy formula for velocity, we get: v = sqrt(2E/m)
04

Substitute the velocity expression back into the radius formula

Substituting the expression for velocity into the equation for the radius, we get: r = m(sqrt(2E/m))/(qB)
05

Simplify the expression for radius

Simplifying the expression, we get: r = sqrt(2mE)/(qB)
06

Ratio of the radii of the electrons' orbits

Let r₁ and r₂ be the radii of the orbits for the electrons with energies E₁ and E₂ respectively. The ratio of the radii is given by: r₁/r₂ = sqrt(2mE₁)/(qB) * (qB)/sqrt(2mE₂)
07

Simplify the ratio expression

Canceling the common terms (qB) and (2m) in the numerator and denominator yields: r₁/r₂ = sqrt(E₁/E₂)
08

Calculate the ratio

Given the energies E₁ = 4.00*10² eV and E₂ = 2.00*10² eV, we can now calculate the ratio of the radii: r₁/r₂ = sqrt((4.00*10²)/(2.00*10²))
09

Simplify the result

Simplifying the result, we get: r₁/r₂ = sqrt(2) So the ratio of the radii of the orbits of the two electrons is √2.

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Most popular questions from this chapter

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.

A square loop of wire of side length \(\ell\) lies in the \(x y\) -plane, with its center at the origin and its sides parallel to the \(x\) - and \(y\) -axes. It carries a current, \(i\), in the counterclockwise direction, as viewed looking down the \(z\) -axis from the positive direction. The loop is in a magnetic field given by \(\vec{B}=\left(B_{0} / a\right)(z \hat{x}+x \hat{z}),\) where \(B_{0}\) is a constant field strength, \(a\) is a constant with the dimension of length, and \(\hat{x}\) and \(\hat{z}\) are unit vectors in the positive \(x\) -direction and positive \(z\) -direction. Calculate the net force on the loop.

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 circular coil with a radius of \(10.0 \mathrm{~cm}\) has 100 turns of wire and carries a current, \(i=100 . \mathrm{mA} .\) It is free to rotate in a region with a constant horizontal magnetic field given by \(\vec{B}=(0.0100 \mathrm{~T}) \hat{x}\). If the unit normal vector to the plane of the coil makes an angle of \(30.0^{\circ}\) with the horizontal, what is the magnitude of the net magnetic torque acting on the coil? 27.61 At \(t=0\) an electron crosses the positive \(y\) -axis (so \(x=0\) ) at \(60.0 \mathrm{~cm}\) from the origin with velocity \(2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. It is in a uniform magnetic field. a) Find the magnitude and the direction of the magnetic field that will cause the electron to cross the \(x\) -axis at \(x=60.0 \mathrm{~cm}\). b) What work is done on the electron during this motion? c) How long will the trip take from \(y\) -axis to \(x\) -axis?

A semicircular loop of wire of radius \(R\) is in the \(x y\) -plane, centered about the origin. The wire carries a current, \(i\), counterclockwise around the semicircle, from \(x=-R\) to \(x=+R\) on the \(x\) -axis. A magnetic field, \(\vec{B}\), is pointing out of the plane, in the positive \(z\) -direction. Calculate the net force on the semicircular loop.
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