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Chapter 27: Problem 21

The magnitude of the magnetic force on a particle with charge \(-2 e\) moving with speed \(v=1.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) is \(3.0 \cdot 10^{-18} \mathrm{~N}\). What is the magnitude of the magnetic field component perpendicular to the direction of motion of the particle?

Short Answer

Expert verified
Answer: The magnitude of the magnetic field component perpendicular to the direction of motion of the particle is approximately \(9.375 \times 10^{-5}\) T.

Step by step solution

01

Write down the given information

We are given that: - the particle has a charge \(q = -2e\), - the speed of the particle is \(v = 1.0 \cdot 10^5\) m/s, - the magnitude of the magnetic force is \(F_B = 3.0 \cdot 10^{-18}\) N.
02

Determine the angle between the velocity vector and the magnetic field vector

In this case, we are looking for the component of the magnetic field that is perpendicular to the direction of motion of the particle, so the angle \(\theta = 90^\circ\).
03

Use the formula for magnetic force to find the magnitude of the magnetic field

The formula for the magnetic force on a moving charged particle is: \(F_B = |q|(vB\sin{\theta})\) Plug in the known values: \(3.0 \cdot 10^{-18} \text{ N} = |-2e|(1.0 \cdot 10^5 \text{ m/s})(B\sin{90^\circ})\) We know that \(e = 1.6 \times 10^{-19} \text{ C}\) and \(\sin{90^\circ} = 1\). Therefore, the equation becomes: \(3.0 \cdot 10^{-18} \text{ N} = |-2(1.6 \times 10^{-19} \text{ C})|(1.0 \cdot 10^5 \text{ m/s})(B)(1)\)
04

Solve for the magnetic field \(B\)

Rearrange the formula to solve for \(B\): \(B = \frac{3.0 \times 10^{-18} \text{ N}}{-2(1.6 \times 10^{-19} \text{ C})(1.0 \times 10^5 \text{ m/s})}\) Now, compute the value of \(B\): \(B \approx 9.375 \times 10^{-5} \text{ T}\)
05

State the final answer

The magnitude of the magnetic field component perpendicular to the direction of motion of the particle is approximately \(9.375 \times 10^{-5}\) T.

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Most popular questions from this chapter

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.

In which direction does a magnetic force act on an electron that is moving in the positive \(x\) -direction in a magnetic field pointing in the positive \(z\) -direction? a) the positive \(y\) -direction b) the negative \(y\) -direction c) the negative \(x\) -direction d) any direction in the \(x y\) -plane

An electron is traveling horizontally from the northwest toward the southeast in a region of space where the Earth's magnetic field is directed horizontally toward the north. What is the direction of the magnetic force on the electron?

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.

A straight wire with a constant current running through it is in Earth's magnetic field, at a location where the magnitude is \(0.43 \mathrm{G}\). What is the minimum current that must flow through the wire for a 10.0 -cm length of it to experience a force of \(1.0 \mathrm{~N} ?\)
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