The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
Speed of the proton = 5.02 × 10 ⁶ m /a
Angel of between the velocity and the magnetic force = 60 °
The magnitude of magnetic field B = 0.180 T
The magnitude of the magnetic force on the proton is,
Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
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The formula is=1/2(m x v^2)
so = 1/2*(0.05)*(310)^2
ans is =2402.5 joules
Te direction of the magnetic force for the velocity of the proton in the
-ve y direction will be +ve z direction.
As we know that the right-hand rule is based on the relation of magnetic fields and the forces that they exert on moving charges.When a charged particle moves under a magnetic field, it exerts a force on the particle, which is not in the same direction but different than the direction of the magnetic field.Under the right-hand rule, if we point our pointer finger in the direction of the charged particle is moving and the middle finger is representing the direction of the magnetic field then our thumb depicts the direction of the magnetic force which is exerted on the charged particle.
So, we are given that the direction of the velocity of the proton is in the negative y direction and the direction of the magnetic field is in the positive x direction, so the magnetic force is acting in the positive z direction.
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Pull the plates apart and you will knwo what it is lmaoo
Answer:
1.1x10^-2N
Explanation:
We have the change in momentum as
P = 0.3(4.5+12)g.mph
= 0.3x0.447x(4.5+12)x10^-3
Then the force that is exerted will be
F = p/∆t
∆t = 0.2
= 0.3x0.447x(4.5+12)x10^-3/0.2
= 0.1341x16.5x10^-3/0.2
= 1.1x10^-2
Therefore the force that was exerted is equal to 1.1x10^-2
