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.
To solve this problem, we can use the formula for the centripetal force in a magnetic field:
F = q(v x B)
where F is the centripetal force, q is the charge of the particle, v is its velocity, and B is the magnetic field.
First, let's find the charge of the particle. The problem states that it is positively charged, so we can assume that it is a proton. The charge of a proton is +1.6 x 10^-19 C.
Next, let's find the radius of the circular path. We know that the particle travels one-quarter of a circle in a time of 1.20 x 10^-3 s. Since the particle moves perpendicular to the magnetic field, its velocity is always tangent to the circular path. Therefore, we can use the formula for the angular velocity of an object in circular motion:
ω = Δθ / Δt
where ω is the angular velocity, Δθ is the change in angle, and Δt is the change in time.
Since the particle moves one-quarter of a circle, the change in angle is 90 degrees or π/2 radians. Plugging in this value, we get:
ω = (π/2) / (1.20 x 10^-3) s
Next, let's find the velocity of the particle. We know that the particle is traveling due east with a speed of 60 m/s. Since the particle enters the magnetic field heading due south, its final velocity will be purely in the south direction. Therefore, the magnitude of its velocity will be 60 m/s. Note that the velocity vector is perpendicular to the magnetic field vector, so the angle between them is 90 degrees.
Now we can use the formula for centripetal force to solve for the magnetic field:
F = q(v x B)
F = mω^2r
where m is the mass of the particle, ω is the angular velocity, and r is the radius of the circular path.
mω^2r = q(v x B)
Solving for B, we get:
B = (mω^2r) / (qv)
Plugging in the given values, we can find the magnetic field B.