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A proton moves in a uniform magnetic fie...

A proton moves in a uniform magnetic field of strength 0.500 T magnetic field is directed along the x-axis. At initial time, t s = 0 , the proton has velocity `vec(v) = (1.95 xx 10^(5)hat(i) + 2.00 xx 10^(5) hat(k)) ms^(-1)`. Find
At initial time, what is the acceleration of the proton.

Text Solution

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Magnetic field `vec(B) = 0.500 hat(i) T`
Velocity of the particle
`vec(v) = (1.95 xx 10^(5)hat(i) + 2.00 xx 10^(5) hat(k))ms^(-1)`
Charge of the proton `q = 1.60 xx 10^(-19)C`.
Mass of the proton `m = 1.67 xx 10^(-27) kg`
The force experienced by the proton is
`vec(F) = q(vec(v) xx vec(B))`
`= 1.60 xx 10^(-19) xx ((1.95 xx 10^(5) hat(i) + 2.00 xx 10^(5) hat(k)) xx (0.500 hat(i)))`
`vec(F) = 1.60 xx 10^(-14) N hat(j)`
Therefore, from Newton’s second law,
`vec(a) = (1)/(m) vec(F) = (1)/(1.67 xx 10^(-27))(1.60 xx 10^(-14))`
`= 9.58 xx 1-^(12) ms^(-2)`
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