A uniform magnetic field of magnitude 0.40 T in the negative z-direction (into t

Question

A uniform magnetic field of magnitude 0.40 T in the negative z-direction (into the screen) is present in a region of space. A uniform electric field is also present. In the figure, the electric field is set at 21, 200 V/m in the positive y-direction. A charge of -3.2 times 10^-19 C is fired through the fields in the positive x-direction. What is the y-component of force on the charge when it has a velocity of 5.3 times 10^4 m/s? +1 times 10^-14 N -7 times 10^-15 N -1 times 10^-14 N -11 times 10^-10 N

Explanation / Answer

Fm = q*VXB

Fe = qE

Net force will be:

Fnet = Fm + Fe

Fnet = q*(VxB) + q*E

Fnet = q*((VxB) + E)

VxB = 5.3*10^4 i x 0.4 (-k) = -2.12*10^4 (ixk) = 2.12*10^4 j

E = 21200 j

Fnet = -3.2*10^-19*(21200 + 21200) j = -1.35*10^-14 N

Fnet = -1*10^-14 N

Correct option is C.

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