An electron (with charge \(-e\) and mass \(m_{\mathrm{e}}\) ) moving in the positive \(x\) -direction enters a velocity selector. The velocity selector consists of crossed electric and magnetic fields: \(\vec{E}\) is directed in the positive \(y\) -direction, and \(\vec{B}\) is directed in the positive \(z\) -direction. For a velocity \(v\) (in the positive \(x\) -direction \(),\) the net force on the electron is zero, and the electron moves straight through the velocity selector. With what velocity will a proton (with charge \(+e\) and mass \(m_{\mathrm{p}}=1836 \mathrm{~m}_{\mathrm{e}}\) ) move straight through the velocity selector? a) \(v\) b) \(-v\) c) \(v / 1836\) d) \(-v / 1836\)

Short Answer

Expert verified
a) v b) -v c) v^2 d) -v^2 Answer: b) -v

Step by step solution

01

Write down the formulas for electric force and magnetic force

The formulas for electric force(F_e) and magnetic force(F_m) are as follows: F_e = q * E F_m = q * v * B Where q is the charge of the particle, E is the electric field, v is the velocity, and B is the magnetic field.
02

Find the relationship between E and B for an electron

When the electron moves straight through the velocity selector, the net force on the electron becomes zero. So we can say: F_e = F_m e * E = e * v_e * B Where v_e is the velocity of the electron. We can cancel out the charge e, and obtain the relationship: E = v_e * B
03

Write the force equations for a proton

Similar to the electron, when the proton moves straight through the selector, the net force on the proton becomes zero. So we can write: F_e = F_m e * E = -e * v_p * B Where v_p is the velocity of the proton. We have a minus sign because a proton has a positive charge, unlike the electron which has a negative charge.
04

Substitute the relationship between E and B from step 2 into step 3 and solve for the proton's velocity

From step 2, we have E = v_e * B. We can plug this into step 3: e * (v_e * B) = -e * v_p * B Notice how we can cancel out the charge e and the magnetic field B: v_e = -v_p Now, we know the mass of the proton is 1836 times the mass of the electron. Therefore, the velocity of the proton (v_p) is: v_p = -v_e = -v So the correct answer is (b) -v.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Force
When studying how particles move through electric and magnetic fields, the electric force plays a critical role. It's one of the fundamental forces that influence charged particles, such as electrons and protons. The strength of the electric force acting on a particle is determined by the equation
\(F_e = q \times E\),
where q represents the charge of the particle and E the electric field intensity. The direction of the force follows the direction of the electric field for a positive charge and is opposite for a negative charge. In a velocity selector, the electric force ensures that only particles with a specific velocity can pass through without deflection, as other velocities would result in the particle being pushed upward or downward by this force, deviating from the straight path.
Magnetic Force
The magnetic force is another component that acts perpendicularly to both the magnetic field and the velocity of a charged particle. This force is fundamentally described by the equation
\(F_m = q \times v \times B\),
where v is the velocity of the particle, B is the magnetic field strength, and q is again the charge. Charged particles moving in a magnetic field experience a force that directs them in a circular or helical path, according to the left-hand rule for negatively charged particles like electrons, and the right-hand rule for positively charged particles like protons. In a velocity selector, the magnetic force must balance exactly with the electric force for a particle to continue moving in a straight line without being deflected to the side.
Charge and Mass of Particles
The charge and mass of particles are two intrinsic properties that greatly influence their motion in electric and magnetic fields. The charge determines the magnitude and direction of the electric and magnetic forces acting on a particle. For instance, a particle with a negative charge like an electron will experience forces in opposite directions when compared to a positively charged particle like a proton when they move through the same electric field.

The mass of a particle, on the other hand, affects its acceleration as a result of these forces according to Newton's second law: \(F = m \times a\). Particles with larger mass will accelerate less under the same force. This is particularly important for velocity selectors, as heavier particles (like protons) need different settings compared to lighter particles (like electrons) to move straight through due to their different mass-to-charge ratios. Thus, while an electron can pass through with velocity \(v\), a proton, which has 1836 times the mass of an electron, would theoretically pass through with a velocity \(v/1836\) if the selection was purely based on mass and charge without considering the existing forces.

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Most popular questions from this chapter

A particle with a charge of \(20.0 \mu \mathrm{C}\) moves along the \(x\) -axis with a speed of \(50.0 \mathrm{~m} / \mathrm{s}\). It enters a magnetic field given by \(\vec{B}=0.300 \hat{y}+0.700 \hat{z},\) in teslas. Determine the magnitude and the direction of the magnetic force on the particle.

In your laboratory, you set up an experiment with an electron gun that emits electrons with energy of \(7.50 \mathrm{keV}\) toward an atomic target. What deflection (magnitude and direction) would Earth's magnetic field \((0.300 \mathrm{G})\) produce in the beam of electrons if the beam is initially directed due east and covers a distance of \(1.00 \mathrm{~m}\) from the gun to the target? (Hint: First calculate the radius of curvature, and then determine how far away from a straight line the electron beam has deviated after \(1.00 \mathrm{~m}\).)

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 current-carrying wire is positioned within a large, uniform magnetic field, \(\vec{B}\). However, the wire experiences no force. Explain how this might be possible.

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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