Mag qs: Acceleration of Particle in Magnetic Field

[suldaman]

Homework Statement

A particle (q = 5.0 nC, m = 3.0 µg) moves in a region where the magnetic field has components Bx = 2.0 mT, By = 3.0 mT, and Bz = -4.0 mT. At an instant when the speed of the particle is 5.0 km/s and the direction of its velocity is 120° relative to the magnetic field, what is the magnitude of the acceleration of the particle?

a. 33 m/s2
b. 17 m/s2
c. 39 m/s2
d. 25 m/s2
e. 45 m/s2

Homework Equations

ma = F = q.v.b.sin(theta)

The Attempt at a Solution

im not entirely sure if i have done this correctly...but here goes.
to make tings easy...
firstly...Bxyz = 5.4mT ?
and...3ug = 3 x 10^(-9) ?

a = ((5nc)(5000)(5.4mT)(sin 120))/(3ug)

pluggin all values gives me...c) 39 m/s2 ?

wat do yall think? have i done anything wrong?

 

[Astronuc]

It seems to be correct with the information given, but it would be helpful if one would write out all the units to make sure that one understands the orders of magnitude (powers of 10).

The force is a vector related to the cross product of the velocity and magnetic field. In 2D, one vector is at a unique angle to another, but in 3D, it is more complicated because the one angle between two vectors lies in a cone.

Normally one would have
$$\\ \vec{F} = det \left( \begin{array} \hat{x} & \hat{y}& \hat{z} \\v_x & v_y & v_z\\B_x & B_y & B_z\end{array}\right)$$

and |v| = $$\sqrt{{v_x}^2\,+\,{v_y}^2\,+\,{v_z}^2}}$$

and similarly for the vector for B.

 

[esalihm]

I would not think I would go and write them all out in this case, makes it too complex and unnecessary.
ur solution look quite right to me