Certain particle of 3 x 10^-19 C is accelerated at 1.6 x 10^13 m/s in a magnetic

Question

Certain particle of 3 x 10^-19 C is accelerated at 1.6 x 10^13 m/s in a magnetic field of 0.5 T, when it enters the field at 3 x 10^6 m/s. Find the mass of the particle. Assume that the angle between velocity and field vectors is 90 degree. Velocity selector of a mass spectrometer use fields of 10 N/C and 0.012 T to set the speed of certain 16x10^-19 C particle. Then the particle enters the second field of 0.5 T. Here it makes an arc of 30 cm. a) Find the speed of the particle as it enters the second field. b) Find the mass of the particle. Formula / Equation / Constant

Explanation / Answer

1.we know that F=qvBsin(theta) here given that theta=90(deg) so F=qvB=ma=>m=qvB/a=2.8125x10^-26 kgs

2.we know E=vB =>10=vx0.012=>v=833.33 m/s

now find the radius of the trajectory of the charge in the new magnetic field and use it in the formula of B2=mE/qRB1 to find the answer.

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