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Chapter 19: Problem 23

An electron moves at speed \(8.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a plane perpendicular to a cyclotron's magnetic field. The magnitude of the magnetic force on the electron is \(1.0 \times 10^{-13} \mathrm{N}\) What is the magnitude of the magnetic field?

Short Answer

Expert verified
Answer: The magnitude of the magnetic field is approximately \(7.81 \times 10^{-5}\,T\).

Step by step solution

01

Write down the given variables

We are given these variables: - Speed of the electron: \(v = 8.0 \times 10^{5}\,\mathrm{m/s}\) - Magnetic force on the electron: \(F = 1.0 \times 10^{-13}\,\mathrm{N}\) - Charge of the electron: \(q = -1.6 \times 10^{-19}\,\mathrm{C}\) - Angle between the velocity vector and the magnetic field vector is \(胃 = 90^{\circ} 鉄� sin(胃) = 1\)
02

Solve for the magnetic field strength

Using the magnetic force formula \(F = qvBsin胃\), we can solve for the magnetic field strength \(B\). Since \(sin(胃) =1\), the equation simplifies to \(F=qvB\). Rearrange the formula to isolate the magnetic field strength: \(B= \cfrac{F}{q \times v}\) Now, plug in the given values: \(B = \cfrac{1.0 \times 10^{-13}\,\mathrm{N}}{(-1.6 \times 10^{-19}\,\mathrm{C) \times (8.0 \times 10^{5}\,\mathrm{m/s})}}\)
03

Calculate the magnetic field strength

Now, calculate the strength of the magnetic field: \(B= \cfrac{1.0\,\times\,10^{-13}\,\mathrm{N}}{((-1.6\,\times\, 10^{-19}\mathrm{C) (8.0\,\times\, 10^{5}\, \mathrm{m/s})}} \approx \boxed{7.81\,\times\,10^{-5}\,T}\) So, the magnitude of the magnetic field is approximately \(7.81 \times 10^{-5}\,T\).

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Most popular questions from this chapter

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.]
A uniform magnetic ficld points north; its magnitude is 1.5 T. A proton with kinetic energy \(8.0 \times 10^{-13} \mathrm{J}\) is moving vertically downward in this field. What is the magnetic force acting on it?
In a certain region of space, there is a uniform electric field \(\overrightarrow{\mathbf{E}}=2.0 \times 10^{4} \mathrm{V} / \mathrm{m}\) to the east and a uniform magnetic field \(\overline{\mathbf{B}}=0.0050 \mathrm{T}\) to the west. (a) What is the electromagnetic force on an electron moving north at \(1.0 \times 10^{7} \mathrm{m} / \mathrm{s} ?(\mathrm{b})\) With the electric and magnetic fields as specified, is there some velocity such that the net electromagnetic force on the electron would be zero? If so, give the magnitude and direction of that velocity. If not, explain briefly why not.
Two parallel wires in a horizontal plane carry currents \(I_{1}\) and \(I_{2}\) to the right. The wires each have length \(L\) and are separated by a distance \(d\). (a) What are the magnitude and direction of the ficld due to wire 1 at the location of wire \(2 ?\) (b) What are the magnitude and direction of the magnetic force on wire 2 due to this field? (c) What are the magnitude and direction of the field due to wire 2 at the location of wire \(1 ?\) (d) What are the magnitude and direction of the magnetic force on wire 1 due to this field? (e) Do parallel currents in the same direction attract or repel? (f) What about parallel currents in opposite directions?
A long straight wire carries a current of 3.2 \(A\) in the positive \(x\)-direction. An electron, traveling at \(6.8 脳 10 6\) m/s in the positive \(x\) -direction, is 4.6 cm from the wire. What force acts on the electron?
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