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natta225 [31]
3 years ago
15

What is the formula of magnetic force ​

Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
5 0

Answer:

F = qvB sinθ

(Hope this helps! Btw, I am the first to answer. Brainliest pls! :D)

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Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
SSSSS [86.1K]

Answer:

{F_{tot} = -1.092*10^{-2}N

Explanation:

The question: What is the net force exerted by these two charges on a third charge q_3 = 47.0 nC placed between q_1 and q_2 at x_3 = -1.240 mm ?

<u>Your answer may be positive or negative, depending on the direction of the force.</u>

Solution:

The coulomb force is given by the equation

F =k\dfrac{q_1 q_2}{d^2}.

where d is the separation between the charges q_1 and q_2.

Now, in our case

q_1=-13.5nC=-13.5*10^{-9}C

q_2=-1.735nC=-1.735*10^{-9}C

q_3= 47*10^{-9}C

The separation between charges q_3 and q_1 is

d_{31}= (-1.240mm)-(-1.735mm)=0.495mm=4.95*10^{-4} m

Therefore, the force between them is

F_{31} = (9*10^9NmC^{-2})\dfrac{47*10^{-9}*(-13.5*10^{-9})}{4.95*10^{-4}} =-0.01152N

and it is directed in the negative x-direction.

The separation between charges q_2  and q_3 is

d_{23} =-1.240mm-0=-1.240*10^{-3}m

therefore, the force between them is

F_{23} =(9*10^9Nm^2C^{-2})\dfrac{47*10^{-9}C*(-1.735*10^{-9}C)}{-1.240*10^{-3}m}=5.94*10^{-4}N

Therefore the total force on charge q_3 is

F_{tot} =5.94*10^{-4}-0.01152N = -0.010926N\\\\\boxed{F_{tot} = -1.092*10^{-2}N}

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How far would a 6 kg ball move in 10 seconds if it is being pushed with a force of 100 N
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