Respuesta :
We are given:
the speed of a proton = 2.0 × 10^5 meters/second;
moving through a magnetic field = 8.5 × 10-2 tesla
direction: east to west
The force and the direction of the force of the proton is determined using the magnetic field formula
F = qvxB
where q = charge
v = vector
x = speed
B = magnetic field
F = 1.6x10^-19 C * 2.0 × 10^5 m/s * 8.5 × 10-^2 tesla
solve for F and the direction of your force is opposite the direction of the proton which is north since the proton is going upwards.
the speed of a proton = 2.0 × 10^5 meters/second;
moving through a magnetic field = 8.5 × 10-2 tesla
direction: east to west
The force and the direction of the force of the proton is determined using the magnetic field formula
F = qvxB
where q = charge
v = vector
x = speed
B = magnetic field
F = 1.6x10^-19 C * 2.0 × 10^5 m/s * 8.5 × 10-^2 tesla
solve for F and the direction of your force is opposite the direction of the proton which is north since the proton is going upwards.
Explanation :
It is given that,
Speed of the proton, [tex]v=2\times 10^{5}\ m/s[/tex]
Magnetic field, [tex]B=8.5\times 10^{-2}\ T[/tex]
Magnetic force is given by :
[tex]F=q(v\times B)[/tex]
q is the charge of proton, [tex]q=1.6\times 10^{-19}\ C[/tex]
So,
[tex]F=1.6\times 10^{-19}\ C\times 2\times 10^{5}\ m/s\times 8.5\times 10^{-2}\ T[/tex]
[tex]F=2.72\times 10^{-15}\ N[/tex]
The direction of force is given by using right hand rule. So, the force will act in upward direction.
Hence, the correct option is (a) " [tex]2.72\times 10^{-15}\ N[/tex] to the north".