Question

Parallel field lines evenly spaced and all pointing right labeled b. There is a charge labeled q₁ with a vector pointing north labeled v₁. There is a second point labeled q₂ with a vector pointing at 55.0 degrees north of east labeled v₂. Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the strength of the magnetic field? T. In the same magnetic field, particle q₂ has a charge of 42.0 μC and a velocity of 1.21 × 10³ m/s. What is the magnitude of the magnetic force exerted on particle 2? N.

Answer 1

The magnetic field strength can be determined using the formula F = qvB sin ϴ, by substituting the given values for the first charge. The magnitude of the magnetic force on the second charge is then calculated using the same formula with the angle specific to its velocity direction.

Step-by-step explanation:

The strength of the magnetic field can be calculated using the formula F = qvB sin ϴ, where F is the magnetic force, q is the charge, v is the velocity of the charged particle, B is the magnetic field strength, and ϴ is the angle between the velocity and the magnetic field. For the first charge, q1, the force, charge, and velocity are given, and we need to solve for B when ϴ is 90 degrees since the charge is moving perpendicular to the magnetic field lines.

The formula rearranges to B = F / (qv sin ϴ). Substituting the given values: B = (5.75 × 10⁻³ N) / (2.7 × 10⁻⁶ C × 773 m/s × sin(90°)), which calculates to a magnetic field strength B.

For the second charge, q2, the same formula is applied, but with the charge's velocity making an angle of 55 degrees with respect to the direction of the magnetic field. The force F2 is then found by plugging in the values for charge q2, its velocity, the calculated magnetic field B, and using sin(55 degrees).