Aparticle with a charge of −1.24×10−8c is moving with instantaneous velocity v⃗ = (4.19×104m/s)i^ + (−3.85×104m/s)j^ . part a what is the force exerted on this particle by a magnetic field b⃗ = (1.80 t ) i^? enter the x, y, and z components of the force separated by commas.

F = 0i (in the x-direction), 0j (in the y-direction),-8.59*10^-4 N k (In the z-direction)

Explanation:

The force given by charged particles moving in a magnetic field is given below (cross is cross product, they don't have that format in the equation tool):

Now we can perform the cross product between v and B

d{array}\right]" alt="v(cross)B = 69300 (kg*m/(s^2*C))\\" />d{array}\right]" />

Now multiply by Q (charge) to get the force

F = -8.59*10^-4 N k

F = 0i, 0j, (-8.59*10^-4)k

6/12=1/2

step-by-step explanation: