An electron moves with a speed of `2 xx 10^5 ms^(-1)` along the positive x-direction in a magnetic field
`barB= (2î – 5hatj – 2.5hatk)` tesla.
The magnitude of the force (in newton) experienced by the electron is found to be `Z xx 10^(-13)`N. What is the numerical value of Z ? (Take `sqrt(5) = 2.24`)
[charge of electron `= 1.6 xx 10^(-19)`C]

Answer

`03.58`

Explanation

`vec(v)=2 xx 10^5 ms^(-1),`
`barF=q(vecV xx barB)`
`=q`
`[(2 xx 10^5 hati) xx (2 hati- 5 hatj-2.5 hatk)]`
`=2 xx 10^5 xx q(-10 hatk + 5 hatj)`
`:. F_z=-20 xx 10^5 xx q`
`:. `Magnitude force
`=sqrt(F_y^2+F_z^2)`
`qsqrt((10^6)^2+(-2 xx 10^6)^2)`
`q xx 10^6 xx 10^(-19) xx 10^6 xx 2.24`
`=3.58 xx 10^(-13)N`