20.08 A particle with a charge of 1.60×10 8 C is moving with an instantaneous ve

Question

20.08

A particle with a charge of 1.60×108 C is moving with an instantaneous velocity of magnitude 45.0 km/s in the x-y plane at an angle of 50.0 counterclockwise from the +x axis.

1.

What is the direction of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the +zdirection?

Enter your answer as a counterclockwise angle from the direction of the force, in degrees, to three significant figures.

2.

What is the magnitude of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the +zdirection?

Enter your answer as a counterclockwise angle from the direction of the force, in degrees, to three significant figures.

Explanation / Answer

given

charge q = - 1.6*10^(-8) C

velocity V = 45 km/s = 45*1000 m/s = 4.5*10^4 m/s

theta = 50 degree

magnetic field B = 2 T

part(1)

velocity in x direction is

Vx = V*cos(theta)

Vx =(4.5*10^(4)) * cos(50)

Vx = 2.89*10^(4) m/s

velocity in y direction

Vy = V*sin(theta)

Vy =(4.5*10^(4)) * sin(50)

Vy = 3.4*10^(4) m/s

force exerted in x direction

Fx = q (Vx*B)

Fx =(1.6*10^(-8))* (( 2.89*10^(4))*2)

Fx = 9.2*10^(-4) N

force exerted in y direction

Fy = q (Vy*B)

Fy =(1.6*10^(-8))* (( 3.4*10^(4))*2)

Fy = 10.88 * 10^(-4) N

Now the direction of force is

tan (theta) = (Fy/Fx)

tan(theta) = ((10.88 * 10^(-4))/(9.2*10^(-4)))

tan(theta) = 1.18

theta = tan^-1(1.18)

theta = 49.72 degree

part(2)

magnitude of force

F = sqrt ((Fx)^2+(Fy)^2)

F= sqrt(( 9.2*10^(-4))^2 +(10.88 * 10^(-4))^2)

F = 14.24 *10^(-4) N

answer

direction of force = 49.72 degree

magnitude of force = 14.24 *10^(-4) N

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