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A particle has 3 x 10^15 eV has a charge of 3uC is placed on a certain field.
2 answers:
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Answer:
, inside the plane
Explanation:
We need to calculate the magnitude and direction of the magnetic field produced by each wire first, using the formula
where
is the vacuum permeability
I is the current
r is the distance from the wire
For the top wire,
I = 4.00 A
r = d/2 = 0.105 m (since we are evaluating the field half-way between the two wires)
so
And using the right-hand rule (thumb in the same direction as the current (to the right), other fingers wrapped around the thumb indicating the direction of the magnetic field lines), we find that the direction of the field lines at point P is inside the plane
For the bottom wire,
I = 5.90 A
r = 0.105 m
so
And using the right-hand rule (thumb in the same direction as the current (to the left), other fingers wrapped around the thumb indicating the direction of the magnetic field lines), we find that the direction of the field lines at point P is also inside the plane
So both field add together at point P, and the magnitude of the resultant field is:
And the direction is inside the plane.
Answer
given,
v = (6 t - 3 t²) m/s
we know,
position of the particle
integrating both side
x = 3 t² - t³
Position of the particle at t= 3 s
x = 3 x 3² - 3³
x = 0 m
now, particle’s deceleration
a = 6 - 6 t
at t= 3 s
a = 6 - 6 x 3
a = -12 m/s²
distance traveled by the particle
x = 3 t² - t³
at t = 0 x = 0
t = 1 s , x = 3 (1)² - 1³ = 2 m
t = 2 s , x = 3(2)² - 2³ = 4 m
t = 3 s , x = 0 m
total distance traveled by the particle
D = distance in 0-1 s + distance in 1 -2 s + distance in 2 -3 s
D = 2 + 4 + 2 = 8 m
average speed of the particle