A positively charged particle of mass 5.60 10-8 kg is traveling due east with a speed of 60 m/s and enters a 0.49-T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 1.20 10-3 s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field.

(a) What is the magnitude of the magnetic force acting on the particle?
1 N

(b) Determine the magnitude of its charge.
2 C

the path of the particle has length PI/2 * radius.

velocity of the particle: PI*radius/2timegiven=60m/s solve for r

Knowing the velocity as that, solve for r.

centripetal force=mv^2/r

then magnetic force, Bqv, solve for that as it equals centripetal force.
then solve for q.

To find the magnitude of the magnetic force acting on the particle, we can use the formula for the magnetic force on a charged particle moving perpendicular to a magnetic field:

F = q v B

Where:
F is the magnetic force,
q is the charge of the particle,
v is the velocity of the particle, and
B is the magnetic field strength.

(a) Given that the particle has a mass of 5.60 × 10^-8 kg, a velocity of 60 m/s, and enters a uniform magnetic field of 0.49 T, we need to find the charge of the particle.

To find the charge of the particle, rearrange the formula as:

q = F / (v B)

Substituting the given values:

q = (m v Δθ) / (r v B)

Where:
q is the charge of the particle,
m is the mass of the particle,
v is the velocity of the particle,
Δθ is the angle through which the particle moves (in this case, 90 degrees as it moves through a quarter-circle),
r is the radius of the circle.

The radius of the circle can be found using the formula for the distance traveled in a circle:

s = r Δθ

Rearranging the formula:

r = s / Δθ

Substituting the given values:

r = (v Δt) / Δθ

Where:
r is the radius of the circle,
v is the velocity of the particle,
Δt is the time it takes for the particle to move through the circle (given as 1.20 × 10^-3 s),
Δθ is the angle through which the particle moves (in this case, 90 degrees as it moves through a quarter-circle).

Now we can substitute this value for r into the formula for q:

q = (m v Δθ) / ((v Δt) / Δθ) B

Simplifying:

q = m v Δθ^2 / (v Δt) B

Substituting the given values:

q = (5.60 × 10^-8 kg) (60 m/s) (90 degrees)^2 / (60 m/s) (1.20 × 10^-3 s) (0.49 T)

Now we can solve for q:

q ≈ 2 C

Therefore, the magnitude of the charge of the particle is approximately 2 C.

(b) We have just found that the magnitude of the charge of the particle is 2 C.