A particle with a charge 7C and a mass of 20kg is traveling in a circular path around a fixed particle of charde -5C. The velocity is observed to be 3000m/sec.
A 9C particle with the same velocity is instead in the presence of a uniform electric field of 250N/C as shown. What is the magnitude and direction of the magnetic field so that it travels in a straight line?
To find the magnitude and direction of the magnetic field required for the 9C particle to travel in a straight line, we can use the formula for the magnetic force on a charged particle moving through a magnetic field.
The magnetic force on a charged particle is given by the formula:
F = q * v * B * sin(θ)
where F is the force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity vector of the particle and the magnetic field vector.
Since we want the particle to travel in a straight line, the magnetic force must be zero. This means that the angle between the velocity vector and the magnetic field vector must be either 0 degrees or 180 degrees.
In this case, the particle with a charge of 9C is in the presence of a uniform electric field, so the magnetic field is present only to change the particle's path. Therefore, we can assume the angle between the velocity vector and the magnetic field vector is 180 degrees (opposite direction).
Given that the charge of the particle is 9C, the velocity is 3000m/s, and we want the force to be zero, we can rearrange the formula to find the magnitude of the magnetic field (B):
B = 0 / (q * v * sin(180°))
sin(180°) is equal to 0, so the magnetic field (B) can be any value since it will not affect the force.
Therefore, the magnitude of the magnetic field can be any value, and the direction does not matter as long as it is opposite to the velocity vector of the particle (180 degrees).