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A particle with a charge of 34?C moves with a speed of 74m/s in the positive x direction. The magnetic field in this region of space has a component of 0.41T in the positive y direction, and a component of 0.86T in the positive zdirection
Part A
What is the magnitude of the magnetic force on the particle?
-->F =_____ mN
Part B
What is the direction of the magnetic force on the particle? (Find the angle measured from the positive z-axis toward the negative y-axis in the yz-plane.)
--> \Theta = _____ \Degrees\Degrees

Added Mon, 03 Aug '15

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Use F = qv (?) B.

Let By and Bz be the magnetic field components in the y-direction and z-direction, respectively.

Let Fy and Fz be the forces caused by By and Bz, respectively.

Since the velocity is normal to both axes, the angle between the vectors when the cross product is computed is 90(?) , and sin(90(?) ) = 1.

Fy = qv (?) By = qvBsin(90(?) )

= (34 (?) C)(74 m/s)(0.41 T)(1).

One tesla = one newton per ampere per meter, and one ampere is one coulomb per second, so rewrite this as

Fy = (34 (?) 10^(-6) C)(74 m/s)(0.41 N/(m(?) C/s)) = 1.0 (?) 10^(-3) N, or 1.0 mN.

Similarly,

Fz = qv (?) Bz = qvBsin(90(?) )

= (34 (?) C)(74 m/s)(0.86 T)(1)

= 2.1 (?) 10^(-3) N, or 2.1 mN.

a) The magnitude of the resultant is ?((1.0)(?) + (2.1)(?) ) = 2.3 (?) 10^(-3) N, or 2.3 mN.

b) Fz is the shorter leg of a right triangle; Fy is the longer leg. (The resultant vector is the hypotenuse.)

The angle ? from the z-axis to the resultant vector is tan?(?) (Fz/Fy)

= tan?(?) (2.1 mN / 1.0 mN)

= 64(?) (approximately)

Added Mon, 03 Aug '15

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