162k views
1 vote
Electron moves in magnetic field having magnitude 12 mT. Find

velocity of the electron when it is exerted a magnetic force 4 fN
and charge of electron is .

User Laniakea
by
7.9k points

1 Answer

0 votes

Final answer:

The question involves calculating the angle between the velocity of an electron and a magnetic field using the magnetic force equation F = qvB sin θ. By rearranging this equation to solve for θ and substituting the given values, we can determine the two possible angles that satisfy the conditions.

Step-by-step explanation:

The student is asking a question related to magnetic forces acting on a moving charge, which is a concept covered in high school physics. The magnetic force exerted on a moving charge in a magnetic field is given by the equation F = qvB sin θ, where F is the magnetic force, q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity of the charge and the direction of the magnetic field.

To find the angle that the velocity of an electron makes with the magnetic field, we can rearrange the formula to solve for θ: θ = arcsin(F / (qvB)). We know the magnetic force F is 1.40 × 10-16 N, the charge q is the charge of an electron (approximately 1.6 × 10-19 C), the velocity v is 4.00 × 103 m/s, and the magnetic field strength B is 1.25 T. Plugging the values into the formula, we can calculate the angle θ. As sinθ can be positive or negative, there will be two possible angles for the direction of the velocity of the electron relative to the magnetic field, symmetrical around the 90-degree axis.

User Grygoriy Gonchar
by
8.4k points