The magnetic field of the Earth at a certain location is...

The magnetic field of the Earth at a certain location is directed vertically downward and has a magnitude of 50.0$\mu \mathrm{T}$ . A proton is moving horizontally toward the west in this field with a speed of $6.20 \times 10^{6} \mathrm{m} / \mathrm{s}$ . (a) What are the direction and magnitude of the magnetic force the field exerts on this particle? (b) What is the radius of the circular arc followed by this proton?

Key Concepts

Lorentz Force

This is the force experienced by a charged particle moving in a magnetic field, mathematically defined as F = q(v × B). The magnitude of the force depends on the charge, its speed, the magnetic field strength, and the sine of the angle between the velocity and the field. It is always perpendicular to both the velocity of the particle and the magnetic field, which is crucial in understanding the subsequent motion of the particle.

Circular Motion in a Magnetic Field

When a charged particle moves perpendicular to a magnetic field, the Lorentz force continuously acts perpendicular to its velocity, causing the particle to execute uniform circular motion. The necessary centripetal force for this motion is supplied entirely by the magnetic force, allowing the derivation of the radius of the circular path from the balance between magnetic force (qvB) and centripetal force (mv²/r), resulting in r = mv/(qB).

Right-Hand Rule

This is a mnemonic used to determine the direction of the force acting on a positively charged particle moving in a magnetic field. By aligning the fingers in the direction of the velocity and curling them towards the magnetic field direction, the thumb points in the direction of the resulting force. This technique is essential for predicting the motion direction of the particle based on the cross product relationship in the Lorentz force.

Magnetic Field

A magnetic field is a vector field that exerts a magnetic force on moving charges and other magnetic dipoles. In the context of Earth, the magnetic field has both magnitude and direction, and its interaction with charged particles is fundamental to many phenomena, including auroras and the behavior of particles in accelerators. Understanding its properties is essential for analyzing the force and motion of charged particles in such environments.

Transcript

00:01 This question concerns a proton moving through earth's magnetic field.

00:05 Using the equation for the force of a charged particle moving through a magnetic field, f equals qv cross b, where f is the force on the particle in newton's, q is the charge of the particle in kulums, v is the velocity vector, and b is the magnetic field vector.

00:32 First, to find the direction, we can use the right -hand rule.

00:36 First, point your thumb in the direction.

00:37 Of the velocity vector.

00:40 Then point the rest of your fingers in the direction of the magnetic field.

00:43 The direction your palm faces is the direction of the force, which in this case would be north.

00:53 The magnitude of the force can be found by the equation f equals qvb sine theta, where theta is the angle between the velocity and magnetic field vectors.

01:08 In this instance, the angle between west which is the direction of the velocity vector, and down, which is the direction of magnetic field vector, is 90 degrees, and the sign of 90 degrees equals 1.

01:27 So, the equation f equals q times v, times b will give us our answer for the magnitude of the force.

01:39 Now, v and b are both given, and since the particle is a proton, the charge of a proton is the elementary charge.

01:47 Which is q equals 1 .602 times 10 to the negative 19th kulams.

01:58 So when we multiply it all together, f equals 1 .602 times 10 to the negative 19th times 6 .2 times 10 to the negative 19th times 6 .2 times 10 to the 6th, times 50 times 10 to the negative 6th, which f then equals 4 .966 times 10 to the negative 17th newton's.

02:35 And going in the north direction...

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