A proton is projected in the positive x direction into a region of uniform electric field E = -5.30*10^5 i N/C at t = 0. The proton travels 6.50 cm as it comes to rest.
(a) Determine the acceleration of the proton.
-5.07E13 (-i) m/s^2
(b) Determine the initial speed of the proton.
______ (+i) m/s
(c) Determine the time interval over which the proton comes to rest.
______ s
A current loop with radius 20 cm and current 2 A is in a uniform magnetic field of 0.5 T. Considering all possible orientations of the loop relative to the field, what is the largest potential energy difference (in Joules) you can find between two orientations?
a proton is projected into a magnetic field that is directed along the positive x-axis. find the direction of the magnetic force exerted on the proton for cach of the following directions of the protons velocity
To solve this problem, we can use the formulas of motion in one dimension under constant acceleration. In this case, the proton experiences a constant acceleration due to the electric field.
(a) To find the acceleration of the proton, we can use the formula:
a = (v - u) / t
where:
a = acceleration
v = final velocity (which is 0 m/s since the proton comes to rest)
u = initial velocity
t = time interval
Here, we are given the final velocity and the initial velocity is unknown. We are also given the time interval, so we can rearrange the formula to solve for acceleration:
a = -u / t (since v = 0)
Plugging in the values:
a = -(-5.30 * 10^5 i N/C) / (6.50 cm) (1 m = 100 cm)
Simplifying the units:
a = 5.30 * 10^5 i N/C / (6.50 * 10^(-2) m)
Calculating the acceleration:
a ≈ 5.07 * 10^13 i m/s^2 (to two significant figures)
Therefore, the acceleration of the proton is approximately -5.07 * 10^13 (-i) m/s^2.
(b) To find the initial speed of the proton, we can use the formula:
v^2 = u^2 + 2as
where:
v = final velocity (which is 0 m/s since the proton comes to rest)
u = initial velocity (unknown)
a = acceleration
s = displacement
Here, we are given the acceleration and displacement. Plugging in the values:
(0)^2 = u^2 + 2 * (5.07 * 10^13 i m/s^2) * (6.50 * 10^(-2) m)
Simplifying:
0 = u^2 + 6.57 * 10^12 i m^2/s^2
Since the initial velocity cannot be complex in this case, we can ignore the imaginary part of the equation. Thus, we have:
0 = u^2
This implies that the initial velocity u = 0 m/s.
Therefore, the initial speed of the proton is 0 (+i) m/s.
(c) To find the time interval over which the proton comes to rest, we can use the formula:
v = u + at
Since v = 0 m/s and u = 0 m/s, the formula becomes:
0 = 0 + (5.07 * 10^13 i m/s^2) * t
Dividing both sides by the acceleration:
0 = 5.07 * 10^13 i m/s^2 * t
Since the acceleration is nonzero, we can conclude that t = 0 s.
Therefore, the time interval over which the proton comes to rest is 0 seconds.