An electron moves with a speed of 1.0 ×10^7 m / s in the directions shown in Figure P24.23. A 0.

An electron moves with a speed of 1.0 ×10^7 m / s in the...

An electron moves with a speed of $1.0 \times 10^{7} \mathrm{m} / \mathrm{s}$ in the directions shown in Figure P24.23. A 0.50 T magnetic field points in the positive $x$ -direction. For each, what is the magnetic force on the electron? Give your answers as a magnitude and a direction.

Key Concepts

Motion of Charged Particles in Magnetic Fields

This concept addresses how charged particles behave when subjected to magnetic forces, including the formation of circular or helical trajectories depending on the orientation of the velocity with respect to the magnetic field. It is also important to note that the magnetic force does no work since it always acts perpendicular to the particle’s velocity.

Right-Hand Rule

This rule is used to determine the direction of the cross product (v × B) in the Lorentz force. For a positive charge, pointing the fingers of the right hand along the velocity and curling them toward the magnetic field gives the direction of the force. For a negative charge, like an electron, the resulting force direction is opposite, which is crucial for accurate prediction of motion.

Lorentz Force

This is the fundamental force experienced by a charged particle moving through a magnetic field, expressed as F = q(v × B). It encapsulates both the magnitude and direction of the force, emphasizing that the force is always perpendicular to both the velocity of the charge and the magnetic field vector.

Cross Product

The cross product operation in the Lorentz force formula determines both the magnitude and direction of the magnetic force. Its magnitude is calculated as |q|vB sin(?), where ? is the angle between the velocity and magnetic field vectors, and its direction is perpendicular to the plane formed by these two vectors.

Transcript

00:01 In the given problem, these are the coordinate axis along horizontal.

00:08 This is the x -axis along vertical.

00:13 This is the z -axis and y -axis is into the plane of paper.

00:21 So this is x, here this is z and this is y -axis.

00:26 Then there is a magnetic field along x -exam.

00:32 And an electron is moving along negative y -axis.

00:42 So this is b and here, this is velocity vector for the electron.

00:49 Now the magnitude of this magnetic field has been given as 0 .50 tesla and the velocity with which the electron is moving is 1 .0 into 10 dash to 7 meter per second in the first part of the problem, the angle between the velocity vector and magnetic field vector is given as 90 degrees.

01:18 So the force acting on the electron means magnetic lawrence force will be given as q means the magnitude of the charge over the electron multiplied by the cross product of v and b, so putting all the known values, 1 .6 into 10 dash par minus 19 kulum for the charge over electron, velocity vector 1 .0 into 10 .7 meter per second, because here it will take a form f equals to qvb, sign of angle between them.

01:58 So we can write it here like 1 .6 into 10 minus 19 gulam multiplied by 1 .0 into 10 dash to the power 7 meters per second multiplied by 0 .50 tesla and for sine 90 this is one only.

02:22 So finally the answer for this force comes out to be 8 .0 into 10 dash to bar minus 13 newton.

02:32 Now we have to find the direction of this force also...

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