A magnetic field of `4xx10^(-3)hatkT` exerts a force of `(4hati+3hatj)xx10^(-10)N` on an particle having a charge of `10^(-9)C` in the XY - plane. Find the velocity of particle.

Explanation

We know that,
`F=q(vxxB)`
Let `v=v_(x)hati+v_(y)hatj+v_(z)hatk`
Now, `vxxB=|(hati,hatj,hatk),(v_(x),v_(y),v_(z)),(0,0,4xx10^(-3))|`
= `4xx10^(-3)v_(y)hati-4xx10^(-3)v_(x)hatj`
So, `F=q(vxxB)`
Here, `F=(4hati+3hatj)xx10^(-10)`,
`q=10^(-9)C`
and `B=4xx10^(-3)hatk=(4hati+3hatj)xx10^(-10)`
= `10^(-9)(4xx10^(-3)v_(y)hati-4xx10^(-3)v_(x)hatj)`
Comparing components of force, we get
`4xx10^(-12)v_(y)=4xx10^(-10)`
`rArrv_(y)=100`
Also, `-4xx10^(-12)v_(x)=3xx10^(-10)`
`rArrv_(x)=-75`
So, `v=v_(x)hati+v_(y)hatj=-75hati+100hatjm//s`