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A charge q=-4.9nC and mass m=18.1picograms is accelerated from the rest through a potential difference of...

A charge q=-4.9nC and mass m=18.1picograms is accelerated from the rest through a potential difference of ΔV=173kV. It enters a region where a uniform B=66mT magnetic field is perpendicular to the velocity of the charge. Determine the radius of the path this charge will follow in the magnetic field( in meters).

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Answer #1

given

q = -4.9*10^-9 C
m = 18.1 pico grams

= 18.1*10^-12 grams

= 18.1*10^-15 kg

delta_V = 173 kV

= 173*10^3 V
B = 66 mT = 0.066 T

kinetic energy gained by the charge particle when it is accelerated,

KE = q*delta_V

we know, KE = p^2/(2*m)

==> p = sqrt(2*m*KE) (where p is momentum)

radius of the path, r = m*v/(B*q)

= p/(B*q)

= sqrt(2*m*KE)/(B*q)

= sqrt(2*18.1*10^-15*4.9*10^-9*173*10^3)/(0.066*4.9*10^-9)

= 17.1 m <<<<<<<<<<<----------------------Answer

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