Question

A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.40 107 m/s and experiences an acceleration of 2.20 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field. magnitude 0.00957 Correct: Your answer is correct. seenKey 0.00957 T direction −y Correct: Your answer is correct. seenKey −y

Answer 1

Answer:

0.00956770833 T

Explanation:

q = Charge of electron = $1.6\times 10^{-19}\ C$

m = Mass of proton = $1.67\times 10^{-27}\ kg$

a = Acceleration = $2.2\times 10^{13}\ m/s^2$

v = Velocity = $2.4\times 10^{7}\ m/s$

Magnetic field is given by

$B=\dfrac{ma}{qv}\\\Rightarrow B=\dfrac{1.67\times 10^{-27}\times 2.2\times 10^{13}}{1.6\times 10^{-19}\times 2.4\times 10^{7}}\\\Rightarrow B=0.00956770833\ T$

The magnetic field is 0.00956770833 T