Magnetic field exist in the space and given as `vecB=-(B_(0))/(l^(2))x^(2)hatk`, where `B_(0)`and `l` positive constants. A particle having positive charge `'q'` and mass `'m'` is pojected wit speed `'v_(0)'` along positive `x` axis from the origin. What is the maimum distance of the charged particle from the `y`-axis before it turns back due to the magnetic field. (Ignore any interaction other than magnetic field)

Options

A. `((ml^(2)v_(0))/(3qB_(0)))^(1//3)`
B. `((3ml^(2)v_(0))/(2qB_(0)))^(1//3)`
C. `((3ml^(2)v_(0))/(qB_(0)))^(1//3)`
D. `((ml^(2)v_(0))/(qB_(0)))^(1//3)`

Answer

C

Explanation

`vecF=qvecvxxvecB`
`=q(v_(x)hati+v_(y)hatj)vecB`
`F_(y)=q (B_(0))/(l^(2)) x^(2)v_(x)`
`int_(0)^(v_(0))dv_(y)=(qB_(0))/(ml^(2))int_(0)^(x_("max"))x^(2) dx`
`v_(0)=(qB_(0))/(ml^(2))(x_("max")^(2))/3`
`x_("max")=((3ml^(2)v_(0))/(qB_(0)))^(1//3)`
Alternate solution
Impulse momentum theorem along `x` -axis
`int_(0)^(x_("max")) (qB_(0)x^(2))/(l^(2))=v_(x)dt=mv_(0)`
`:.int_(0)^(x_("max"))(qB_(0)x^(2))/(l^(2))=dx=mv_(0)`
`:.x_("max")=((3ml^(2)v_(0))/(qB_(0)))^(1//3)`