A particle (m=3*10^-6 g, q=5*10^-6 C) moves in a uniform magnetic field given by (60j) mT. At t=0 the velocity of the particle is equal to (30j - 40k)m/s. The subsequent path of the particle is?
I know that the answer is: helical with a 40-cm radius. But i don't know the solution to get to this answer. I really need help. Thanks for anything!
To find the subsequent path of the particle in a uniform magnetic field, we can use the Lorentz force equation:
F = q(v x B)
Where,
F = Force experienced by the particle
q = Charge of the particle
v = Velocity vector of the particle
B = Magnetic field vector
Let's calculate the force first:
F = q(v x B)
Given that q = 5 * 10^-6 C and B = 60j mT = 60 * 10^-3 T,
F = 5 * 10^-6 C * (30j - 40k)m/s x (60 * 10^-3 T)j
F = (5 * 30 * 60 * 10^-9 N)j x j - (5 * 40 * 60 * 10^-9 N)j x k
F = (9 * 10^-8 N)j - (12 * 10^-8 N)k
Next, we calculate the acceleration:
a = F / m
Given that the mass of the particle is m = 3 * 10^-6 g = 3 * 10^-9 kg,
a = (9 * 10^-8 N)j - (12 * 10^-8 N)k / (3 * 10^-9 kg)
a = (3 * 10^1 m/s^2)j - (4 * 10^1 m/s^2)k
Now, we find the radius of the helical path using the centripetal acceleration formula:
a = v^2 / r
Given that the velocity v = 30j - 40k m/s,
(3 * 10^1 m/s^2)j - (4 * 10^1 m/s^2)k = (30j - 40k m/s)^2 / r
(3 * 10^1 m/s^2)j - (4 * 10^1 m/s^2)k = (900j^2 - 2400jk + 1600k^2) m^2/s^2 / r
Equating the coefficients of respective vectors,
30 = 900 / r (coefficient of j)
40 = 1600 / r (coefficient of k)
From the second equation, we can find the value of r as:
r = 40 * 1600 / 1600
r = 40 m
The subsequent path of the particle is helical with a radius of 40 cm.