A non-uniform magnetic field `vecB=B_(0)(1+ y/d)(-hatk)` is present in the region of space between `y=0` and `y=d`. A particle of mass `m` and positive charge `q` has velocity `vecv=(3qB_(0)d)/mhati` at origin `O`. Find the angle made by velocity of the particle withe the positive `x`-axis when it leaves the field. (Ignore any interaction other than magnetic field)

Options

A. `90^(@)`
B. `30^(@)`
C. `60^(@)`
D. `45^(@)`

Answer

C

Explanation

`vecF=q(v_(x)hati+v_(y)hatj)xxB_(0)(1+y/d)(-hatk)=F_(x)hati+F_(y)hatj`
`|F_(x)|=qB_(0)(1+y/d)v_(y)`

`-int_(v_(0))^(v_(x))mdv_(x)=qint_(0)^(d)B_(0)(1+y/d) dy ((dv_(x))/(dy) "is negative")`
`v_(x)=v_(0)-(qB_(0))/m(3/2d)`
`v_(x)=(3qB_(0)d)/m-3/2 (qB_(o)d)/m=3/2(qB_(0)d)/m=(v_(0))/2`
`v_(x)^(2)+v_(y)^(2)=v_(0)^(2)`
`(v_(0)^(2))/4+v_(y)^(2)=v_(0)^(2)`
`impliesv_(y)=(sqrt(3v_(0)))/2`
`tantheta=(v_(y))/(v_(x))=sqrt(3)`
`theta=60^(@)`