A particle that has a charge of 7.6 μC moves with a velocity of magnitude 4...

A particle that has a charge of 7.6 μC moves with a velocity of magnitude 4 × 105 m/s along the +x axis. It experiences no magnetic force, although there is a magnetic field present. The maximum possible magnetic force that the charge with the given speed could experience has a magnitude of 0.310 N. Find the magnitude and direction of the magnetic field. Note that there are two possible answers for the direction of the field.

Solutions

Expert Solution

According to Lorentz force equation, a charged particle of charge q moving with a velocity v in a magnetic field B will experience a magnetic force F=qvBsinθ.. Here θ is the angle between the magnetic field direction and direction of the velocity of the particle. So from the above equation we can see that the magnitude of the magnetic force depends on the angle between the magnetic field and velocity

Given-

Magnitude of the electric charge available on the particle q =7.6×10^-6 C

Velocity of the charged particle along the positive x axis vx=4×10⁵ m/s

In the presence of magnetic field the charged particle experiences no force

The magnetic field is strong enough to exert a maximum force F=0.310 N

Let B be the magnetic field existing in the region.

Then the maximum magnetic force can be expressed as F=qvB

So the magnitude of the magnetic field

B=F/qv

It is mentioned in the question that the charged particle experiences no force.

We have the equation for magnetic force F=qvBsin⁡θ

Here the terms q, v and B are not equal to zero

Then the force F can be zero with the value of the sin function turning to zero.

The term sinθ=0 for θ=0 or θ=180

That is the magnetic field may be either parallel to the direction of motion or anti parallel to the direction of motion.

genius_generous answered 3 weeks ago

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