Question

The magnetic field in a region is given by B = (0.750 + 0.270) T. At some instant, a particle with charge q = 28.0 mC has velocity v = (35.6 + 107.3 + 44.5) m/s. What is the magnetic force exerted on the particle at that instant? (Express your answer in vector form.)

Answer 1

F = (-0.3375i + 0.875j - 1.832k)

Step-by-step explanation:

Q = 28mC = 28 * 10^-3 C

V = 35.6i + 107.3j + 44.5k

B = 0.750i + 0.270j

F = qv * B (cross product vectors)

qv = 28*10⁻³ * (35.6i + 107.3j + 44.5k)

qv = (0.996i + 3.0j + 1.25k)

The force on a charge is the cross product of the charge-velocity and Magnetic field force on the charge.

F= Qv * B

F = (0.996i + 3.0j + 1.25k) x (0.70i + 0.270j)

Resolve the above vectors using 3x2 matrix method. (See attached photo for clarity)

F = (-0.3375i + 0.875j - 1.832k)

Answer 2

Answer: F = (-0.33642, 0.9345,- 1.984164) N

Step-by-step explanation:

The magnetic force can be calculated as:

F = q*(vxB)

v = (vx, vy, vz) = (35.6, 107.3, 44.5) m/s

B = (Bx, By, Bz) = (0.750, 0.270, 0 ) T

where the cross product can be calculated as:

the cross product is:

vxB = ((vy*Bz - vz*By), (vz*Bx - vx*Bz), (vx*By - vy*Bx))

now, we can replace the values and get:

vxB = (107.3*0 - 44.5*0.270), (44.5*0.750 - 0), ( 35.6*0.270 - 0.750*107.3)) T*m/s

= (-12.015, 33.375, -70.863) T*m/s

Then the magnetic force is:

F = (28.0)mC * (-12.015, 33.375, -70.863) T*m/s

remember that the unit of Teslas is in coulombs, so we must writ 28mC as

F = (0.028)*C* (-12.015, 33.375, -70.863) T*m/s

F = (-0.33642, 0.9345,- 1.984164) N