A particle with mass 1.45×10 3 kg and charge of +1.37×10 8 C has, at a given ins

Q: A particle with mass 1.45×10 3 kg and charge of +1.37×10 8 C has, at a given ins

magnetic force F = q*( v x B ) v = 3.2*10^4 j B = 1.13*cos45i + 1.13*sin45 j B = 0.799 i + 0.799 j F = 1.37*10^-8*( 3.2*10^4 j x (0.799 i + 0.799 j) ) F = 1.37*10^-8*3.2*10^4*0.799 -k N F = -0.0003503 k N acceleration a = F/m a = 0.0003503/(1.45*10^-3) a = 0.242 m/s^2

ID: 1609408 • Letter: A

Question

A particle with mass 1.45×103 kg and charge of +1.37×108 C has, at a given instant, a velocity of 3.10×104 m/s along the +y-axis, as shown in the figure. (Figure 1)

Part A

What is the magnitude of the particle’s acceleration produced by a magnetic field of magnitude 1.13 T in the xy plane, directed at an angle of 45.0 counterclockwise from the +x-axis?

Express your answer with the appropriate units to three significant figures.

Part B:

What is the direction of the particle’s acceleration?

45.00 C

Explanation / Answer

magnetic force F = q*( v x B )

v = 3.2*10^4 j

B = 1.13*cos45i + 1.13*sin45 j

B = 0.799 i + 0.799 j

F = 1.37*10^-8*( 3.2*10^4 j x (0.799 i + 0.799 j) )

F = 1.37*10^-8*3.2*10^4*0.799 -k N

F = -0.0003503 k N

acceleration a = F/m

a = 0.0003503/(1.45*10^-3)

a = 0.242 m/s^2 <<<<<<-------answer

part B

direction = -z axis <<<<<<------answer

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A particle with mass 1.45×10 3 kg and charge of +1.37×10 8 C has, at a given ins

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