The magnetic force experienced by the proton is given by
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and
the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so
and we can ignore it in the formula.
For Netwon's second law, the force is also equal to the proton mass times its acceleration:
So we have
from which we can find the magnitude of the field:
Answer:I believe it is D I might be wrong
Explanation:
Absolutely false!! Creativity, curiosity encouraged in Scientific endeavors. Science is all about exploring knowledge, it just promotes intelligency!
In Short, Your Answer would be "False"
Hope this helps!
Answer:
à in unit vector notation = 12.26485i + 7.54539j
B in unit vector notation = 16.3516i + 3.11529j
Explanation:
The detailed steps and calculation is shown in the attachment.
Answer:
A. -2.16 * 10^(-5) N
B. 9 * 10^(-7) N
Explanation:
Parameters given:
Distance between their centres, r = 0.3 m
Charge in first sphere, Q1 = 12 * 10^(-9) C
Charge in second sphere, Q2 = -18 * 10^(-9) C
A. Electrostatic force exerted on one sphere by the other is:
F = (k * Q1 * Q2) / r²
F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²
F = -2.16 * 10^(-5) N
B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:
Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))
= - 6 * 10^(-9) C
Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C
Hence the electrostatic force between them is:
F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²
F = 9 * 10^(-7) N
