A proton (q=1.60 ×10^-19 C, m=1.67 ×10^-27 kg) moves in a uniform magnetic field B=(0.500 T) î .

A proton (q=1.60 ×10^-19 C, m=1.67 ×10^-27 kg) moves in a...

A proton $\left(q=1.60 \times 10^{-19} \mathrm{C}, m=1.67 \times 10^{-27} \mathrm{kg}\right)$ moves in a uniform magnetic field $\overrightarrow{\boldsymbol{B}}=(0.500 \mathrm{T}) \hat{\boldsymbol{i}} .$ At $t=0$ the proton has velocity components $v_{x}=1.50 \times 10^{5} \mathrm{m} / \mathrm{s}, v_{y}=0,$ and $v_{z}=2.00 \times 10^{5} \mathrm{m} / \mathrm{s}(\text { see Example } 27.4) .$ (a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric field in the $+x$ -direction, $\vec{E}=\left(+2.00 \times 10^{4} \mathrm{V} / \mathrm{m}\right) \hat{\imath}$ (b) Will the proton have a component of acceleration in the direction of the electric field?(c) Describe the path of the proton. Does the electric field affect the radius of the helix? Explain. (d) At $t=T / 2$ , where $T$ is the period of the circular motion of the proton, what is the $x$ -component of the displacement of the proton from its position at $t=0 ?$

Key Concepts

Lorentz Force

The Lorentz force is the total force acting on a charged particle moving in electric and magnetic fields. It is given by the vector sum F = q(E + v × B), where q is the particle’s charge, E is the electric field, v is the velocity vector, and B is the magnetic field. This law is essential for analyzing how charged particles are accelerated and deflected by fields.

Magnetic Force on Charged Particles

When a charged particle moves in a magnetic field, it experiences a force that is perpendicular to both its velocity and the magnetic field. This force, given by F = q v × B, does not do work on the particle because it is always perpendicular to the particle's instantaneous velocity. The resulting circular or helical motion is a fundamental aspect of charged particle dynamics in magnetic fields.

Electric Force on Charged Particles

In a uniform electric field, a charged particle experiences a constant force in the direction of the field (or opposite, depending on the sign of the charge). This force, F = qE, results in linear acceleration along the field direction, affecting the particle's speed and trajectory. Understanding the electric force is crucial for predicting how the particle’s motion is modified when electric fields are present.

Circular Motion in a Magnetic Field

A charged particle moving perpendicular to a uniform magnetic field undergoes circular motion with a radius determined by balancing the magnetic force with the centripetal force. The period of this circular motion, often called the cyclotron period, depends on the particle's mass, charge, and the magnetic field strength. This concept is key in many applications, including cyclotrons and mass spectrometers.

Helical Motion

When a charged particle has velocity components both perpendicular and parallel to a uniform magnetic field, its trajectory becomes a helix. The perpendicular component produces circular motion, while the parallel component results in uniform linear movement along the field lines. The overall path is thus a helix, and the separation of these components is fundamental in understanding the combined effects of electromagnetic forces on particle trajectories.

Transcript

00:03 Well, consider a proton whose charge q is equal to 1 .60 times 10 to the power minus 19 column.

00:16 And mass of proton, well, mass of proton is equal to 1 .67 times 10 to the power minus 27 kilogram.

00:27 The proton moves in a uniform magnetic field and magnetic field is equal to 0 .5000 tesla and it is in the i -cam direction.

00:48 8 times t is equal to 0.

00:51 It time t is equal to 0.

00:53 The proton has a velocity v which is equal to 1 .5.

01:02 0 times 10 to the power 5 meter per second that's in the i -cap direction that's one component of velocity plus 2 .00 times 10 to the power 5 meter per second that is the second component and it is in the k -cap direction right well this is vx velocity along x and this is v z velocity along z axis now the force is given by well the force f is equal to q into e factor that's electric field plus v cross b which is equal to q into b cross b now this can the further written is f is equal to q into vxi cap, its velocity along x -axis, plus vz k -cap, its velocity along z -axis, multiply by bx -i -cap, right? and this is equal to q times vz times bxj cap.

03:01 Now let's do substitution.

03:05 Well, by doing substitution, we have force f is equal to 1 .60 times 10 to the power minus 19 column into 2 .00 times 10 to the power 5 meter per second into 0 .50 tesla.

03:35 Cap direction so force vector is equal to 1 .60 times 10 to the power minus 14 newton in the jcap direction now let's do part b now we do part b well we have an electric field e which is equal to plus 2 .00 times 10 to the power 4 volt per meter and its direction is along i cap.

04:18 This field will exert a force on the proton in the same direction of the field...

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