An electron has a velocity of 1.00 km/s (in the positive x direction) and an acc

Question

An electron has a velocity of 1.00 km/s (in the positive x direction) and an acceleration of 2.00 x 1012 m/s2 (in the positive z direction) in uniform electric and magnetic fields. If the electric field has a magnitude of strength of 25.0 N/C (in the positive-ve z direction), determine the following components of the magnetic field. If a component cannot be determined, enter undetermined'. Bx = undetermined By = 0.0113 Your response differs from the correct answer by more than 10%. Double check your calculation. Bz = 0 The proton moves with a velocity of = (5i^ - 4j^ + k^) m/s in a region in which the magnetic field is = (i^ + 2j^ - 3k^)T. What is the magnitude force this charge experiences? N

Explanation / Answer

magnetic field and electric field are related by E = Bc

By = 25 /1000    = 0.025 T J

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magnetic force of charged particle palced in uniorm magnteic field is given by

F = qVB sin theta

where i is current

q is is lcharge

v is speed

B = magnetic field

theta is the angle betweeen L and B

so here we apply

consider ( V x B ) =        i       j        k

                                          5     -4       1

                                          1      2      -3

v x B = i (12 -2) - j(-15-1)    + k(10+4)

v x B = 10 i + 16 j + 14 k

so

F = Fxi + Fyj + Fz k = 1.6 e-19*(10 i + 16 j + 14 k )

F = 1.6 e -18 i    + 2.56 e -18 j     + 2.24 e -18 k

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