A proton of mass $1.67 \times 1 0^{- 27}$ kg and charge $1.68 \times 1 0^{- 19}$ C is projected with a speed of $2 \times 1 0^{6} m s^{- 1}$ at an angle of $6 0^{o}$ to the $X -$ axis. If a uniform magnetic field of $0.10$ T is applied along $Y -$ axis, the path of proton is :

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Answer 1

$v_{x} = v cos θ$
$v_{y} = v sin θ$
Since $v_{y} \neq = 0$ , this component will be responsible for motion along the direction $B ($ i.e $y -$ axis $)$ .
The path will be helical and pitch will be constant.
Radius $r$ of the circle perpendicular to the magnetic field $B$

$r = \frac{m v cos θ}{q B} = \frac{1 . 6 \times 1 0 ^{- 27} \times ( 2 \times 1 0 ^{6} \times cos 6 0 ^{0} )}{1 . 6 \times 1 0 ^{- 19} \times 0 . 1} = 0.1 m$

$T = \frac{2 π r}{v} = \frac{2 \times π \times 0 . 1}{2 \times 1 0 ^{6} \times cos 6 0 ^{0}} = 2 π \times 1 0^{- 7} s e c s$

Answer 2

v_{x} = v cos θ

v_{y} = v sin θ

Since

v_{y} \neq = 0

, this component will be responsible for motion along the direction

B (

i.e

y -

axis

)

.
The path will be helical and pitch will be constant.
Radius

r

of the circle perpendicular to the magnetic field

B

r = \frac{m v cos θ}{q B} = \frac{1 . 6 \times 1 0 ^{- 27} \times ( 2 \times 1 0 ^{6} \times cos 6 0 ^{0} )}{1 . 6 \times 1 0 ^{- 19} \times 0 . 1} = 0.1 m

T = \frac{2 π r}{v} = \frac{2 \times π \times 0 . 1}{2 \times 1 0 ^{6} \times cos 6 0 ^{0}} = 2 π \times 1 0^{- 7} s e c s