Question

Magnetic field question, please help?

A charge q=−0.2μC has a velocity of 1500000m/s at 45∘ to the +x axis in the xz plane, with a positive x and z component. There is a magnetic field of magnitude 0.03T.

a) If B is along the +zaxis, what is the force on the charge?

_____ N along the +y axis

b) If the force on the charge is 2×10^−3 N along +y axis, what is the direction of B?

_____ degrees from the +z axis in the xz plane.

Answer 1

This is what I got

Explanation:

Force experienced by a charge `q` moving with velocity `vecv`in a magnetic field `vecB` is given by part expression of Lorentz Force equation

`vecF_B=q(vecvxxvecB)`

(a) Magnetic field is along `hatz`. Velocity has components along `hatxand hatz`. We see that in the cross product parallel to `hat z` component of velocity vanishes. Hence we get

`vecF_B=(-0.2xx10^-6)(1500000\ cos 45^@hatx xx0.03hatz)`
`=>vecF_B=0.00473\ haty\ N`

(b) Taking the reverse route. Assuming magnitude of the velocity remains same and `x`-component of velocity makes an angle `theta` with this axis

`2×10^-3haty=(-0.2xx10^-6)(1500000\ cos theta^@hatx xx0.03hatz)`
`=>1=(0.1xx10^-3)(1500000\ cos theta^@ xx0.03)`
`=>cos theta^@ =1/((0.1xx10^-3)(1500000xx0.03))`
`=> theta =1.3^@`
`:.`Direction of `B` makes an angle `90-1.3=86.7^@` from the `+z`-axis in the `xz` plane.