You set up a uniform magnetic field with a strength of 0.10 T, pointing toward t

Question

You set up a uniform magnetic field with a strength of 0.10 T, pointing toward the positive x-direction in your lab. You release a particle of charge 11.25 C into the magnetic field region. It travels velocity 250i^+180j^+170k^ in units of m/s. Assuming that your instrument readings are working properly, what are the components of the magnetic force at the moment you release the particle? [Enter the x-component in the first box, the y-component in the second, and the z-component in the third]

The particle's motion is helical, moving forward parallel to the x-axis while turning in circles parallel to the yz-plane. The radius of the particle's circular motion is 50 cm. You want to figure out the particle's mass

Explanation / Answer

Fb = q*( v x B )

v x B = (250i + 180j + 170 k) x (0.1 i)

v x B = (180*0.1)(-k) + (170*0.1)(-j)

v x B = -18k - 17 j

Fb = q*( vx B

Fb = -11.25*10^-6*17 j - 11.25*10^-6*18 k

Fb = -191.25*10^-6 j - 202.5*10^-6 k

Fb = ( 0 , -191.25*10^-6   ,   202.5*10^-6)

part(B)

in y z plane

r = m*vx/qB

0.5 = (m*250)/(11.25*10^-6*0.1)

m = 2.250*10^-9 kg

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