A particle with a charge of 1.85×108 C is moving with an instantaneous velocity

Question

A particle with a charge of 1.85×108 C is moving with an instantaneous velocity of magnitude 40.5 km/s in the xy -plane at an angle of 50.5 o counterclockwise from the +x axis.

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

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

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

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

Explanation / Answer

q = -1.85 * 10^-8 C
v = 40.5 * 10^3 m/s

(a)
Direction of the force exerted on particle, Perpendicular to VB plane, towards + Z axis.
(b)
Magnitude of the force exerted on particle -
Magnetic Force on a moving charge = q*V X B.
Magnetic Force on a moving charge = q*V.B * sin(theta).
Magnetic Force = 1.85 * 10^-8 *40.5 * 10^3* 2 * sin(180-50.5)
Magnetic Force = 1.15 * 10^-3 N

(C)
Direction of the force exerted on particle, Perpendicular to the Direction of motion of Particle towards -ve Y axis.
(D)
Magnitude of the force exerted on particle -
theta = 90o , as Magnetic Field is perependicular to velocity plane.

Magnetic Force on a moving charge = q*V X B.
Magnetic Force on a moving charge = q*V.B * sin(theta).
Magnetic Force = 1.85 * 10^-8 *40.5 * 10^3* 2 * sin(90)
Magnetic Force = 1.5 * 10^-3 N

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