Respuesta :
Answer: F = (-0.33642, 0.9345,- 1.984164) N
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
Answer:
F = (-0.3375i + 0.875j - 1.832k)
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)
