Question

An electron moves through a uniform magnetic field given by = Bx+(3.39 Bx). At a particular instant, the electron has velocity = (2.59+4.09) m/s and the magnetic force acting on it is (4.22 × 10-19) N. Find Bx.

Answer 1

`B_x=-0.56 T`

Explanation:

We are given that an electron moves through a uniform magnetic field given by

`B=B_x\hat{i}+3..39 B_x\hat{j}`

Velocity of electron=
`2.59\hat{i}+4.09\hat{j}`m/s

Magnetic force acting on electron, F=
`4.22* 10^(-19) N`

We have to find
`B_x`

Charge on electron=q=
`-1.6* 10^(-19)C`

We know that magnetic force on electron is given by

`F=q\vec{v}* \vec{B}`

Substitute the value then we get

`4.22* 10^(-19)=-1.6* 10^(-19)(2.59\cdot 3.39B_x-4.09B_x)`

`(8.7801B_x-4.09B_x)=(4.22* 10^(-19))/(-1.6* 10^(-19))`

`4.6901B_x=-(4.22)/(1.6)`

`B_x=-(4.22)/(1.6* 4.6901)`

`B_x=-0.56 T`