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Chapter 19: Problem 11

A uniform magnetic field points vertically upward; its magnitude is $0.800 \mathrm{T}\(. An electron with kinetic energy \)7.2 \times 10^{-18} \mathrm{J}$ is moving horizontally eastward in this field. What is the magnetic force acting on it?

Short Answer

Expert verified
Answer: To find the magnetic force, first, determine the electron's velocity using its kinetic energy, mass, and the equation \(v = \sqrt{\frac{2*KE}{m}}\). Then, calculate the magnetic force using the equation F = q(v x B), where F is the magnetic force, q is the charge of the electron, v is its velocity, and B is the magnetic field. After plugging in the values for q, v, and B and solving the expression, you will find the magnetic force acting on the electron.

Step by step solution

01

Find the electron's velocity

First, we will find the velocity of the electron using its kinetic energy and mass. The kinetic energy (KE) of the electron can be given by: KE = \(\frac{1}{2}mv^2\), where m is the mass of the electron and v is its velocity. Rearranging the equation for v, we get: v = \(\sqrt{\frac{2*KE}{m}}\) The mass of an electron (m) is approximately \(9.11 * 10^{-31} kg\) and the given kinetic energy (KE) is \(7.2 * 10^{-18} J\). Plugging these values into the equation, we can find the electron's velocity: v = \(\sqrt{\frac{2*(7.2*10^{-18})}{(9.11*10^{-31})}}\)
02

Calculate the magnetic force

The magnetic force acting on a charged particle is given by the following equation: F = q(v x B) Here, F is the magnetic force, q is the charge of the particle, v is its velocity, and B is the magnetic field. Since the electron is moving horizontally eastward in a vertically upward magnetic field, the angle between the velocity vector and magnetic field vector is 90 degrees. Therefore, the cross product (v x B) equals the product of v and B: v x B = v * B Now, we can plug in the values to find the magnetic force: F = (-1.6 * 10^{-19})(v)(0.800) (Note that the charge of an electron is negative, so we use -1.6 * 10^{-19} C for the charge.)
03

Compute the magnetic force

Now that we have found the electron's velocity and have the equation for magnetic force, we can plug in the values to compute the magnetic force acting on the electron: F = (-1.6 * 10^{-19})(v)(0.800) Using the value of v from Step 1, compute the final magnetic force acting on the electron: F = (-1.6 * 10^{-19})(\(\sqrt{\frac{2*(7.2*10^{-18})}{(9.11*10^{-31})}}\))(0.800) After evaluating this expression, the magnetic force acting on the electron can be determined.

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Most popular questions from this chapter

A strip of copper \(2.0 \mathrm{cm}\) wide carries a current \(I=\) $30.0 \mathrm{A}\( to the right. The strip is in a magnetic field \)B=5.0 \mathrm{T}$ into the page. (a) What is the direction of the average magnetic force on the conduction electrons? (b) The Hall voltage is \(20.0 \mu \mathrm{V}\). What is the drift velocity?
Two long straight wires carry the same amount of current in the directions indicated. The wires cross each other in the plane of the paper. Rank points \(A, B\) \(C,\) and \(D\) in order of decreasing field strength.
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.
At a certain point on Earth's surface in the southern hemisphere, the magnetic field has a magnitude of \(5.0 \times 10^{-5} \mathrm{T}\) and points upward and toward the north at an angle of \(55^{\circ}\) above the horizontal. A cosmic ray muon with the same charge as an electron and a mass of $1.9 \times 10^{-28} \mathrm{kg}$ is moving directly down toward Earth's surface with a speed of \(4.5 \times 10^{7} \mathrm{m} / \mathrm{s} .\) What is the magnitude and direction of the force on the muon?
An electromagnet is made by inserting a soft iron core into a solenoid. The solenoid has 1800 turns, radius \(2.0 \mathrm{cm},\) and length $15 \mathrm{cm} .\( When \)2.0 \mathrm{A}$ of current flows through the solenoid, the magnetic field inside the iron core has magnitude 0.42 T. What is the relative permeability \(\overline{K_{\mathrm{B}}}\) of the iron core? (See Section 19.10 for the definition of \(\boldsymbol{K}_{\mathbf{B}} .\) )
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