A magnetic field of `(4.0 X10^(-3)vec k)` T exerts a force of `(4.0veci +3.0vecj )X 10^(-10)` N on a particle having a charge of `1.0X10^(-9)` C and going in the x-y plane. Find the velocity of the particle.

Explanation

Magnetic force `vecF_(m)=(4.0hati+3.0hatj)xx10^(-10)N`
Let velocity of the particle in x-y plane be,`vecv=v_(x)hati+v_(y)hatj`
Then from the relation `vecF_(m)=1(vecvxxvecB)`
We have `(4.0hati+3.0hatj)xx10^(-10)=10^(-9)(v_(x)hati+v_(y)hatj)xx(4xx10^(-3)hatk)]=(4v_(y)xx10^(-12)hati-4v_(x)xx10^(-12)hatj)` . comparing the coefficient of `hati` and `hatj` we have,
`4xx10^(-10)=4v_(y)xx10^(12)`
`thereforev_(y)=10^(2)m//s=100m//s` and `3.0xx10^(-10)=-4v_(x)xx10^(-12)`
`thereforev_(x)=-75m//sthereforevecv=-75hati+100hatj`