Question

A charged particle carrying charge `q=1muc` moves in uniform magnetic with velocity `v_1=10^6m//s` at angle `45^@` with `x`-axis in the `xy`-plane and experiences a force `F_1=5sqrt2mN` along the negative `z`-axis. When te same particle moves with velocity `v_2=10^6m//s` along the `z`-axis it experiences a force `F_2` in `y`-direction. Find
a. the magnitude and direction of the magnetic field
b. the magnetic of the force `F_2`.

Answer 1

`F_2` is in `y`-direction when velocity is along `z`-axis. Therfore, magnetic field should be along `x`-axis. So let,
`B=B_0hati`
a. Given, `v_1=10^6/sqrt2hati+10^6/sqrt2hatj`
and `F_1=-5sqrt2xx10^-3hatk`
From the equation `F=q(vxxB)`
We have `(-5sqrt2xx10^-3)hatk=(10^-6)[(10^6/sqrt2hati+10^6/sqrt2hatj)xx(B_0hati)]`
`=-B_0/sqrt2hatk`
`:. B_0/sqrt2=5sqrt2xx10^-3`
or `B_0=10^-2T`
Therefore, the magnetic field is
`B=(10^-2hati)T`
b. `F_2=B_0qv_2sin90^@`
As the angle between `B` and `v` in this case is `90@`
`:. F_2=(10^-2)(10^-6)(10^6)`
`=10^-2N`