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.