An electrostatic force of 5 N exists between two identical point charges, each of mass 0.01 g, placed 4 mm apart.
a.) Find the magnitude of each charge.
b.) One of these charges is placed into a magnetic field. Assuming that the charge has a speed of 50 m/s and travels a circular path of radius 5 mm, what is the strength of the magnetic field?
c.) Now the charge is removed and a 0.5 m long wire is placed into the magnetic field from part b. What magnitude current would have to pass through the wire in order for it to feel the same magnetic force as the charges in part a?
(a) Use the Coulomb force formula for the force acting on each of two charges.
F = k Q1 Q2/R^2
In your case Q1 = Q2 = Q, and R = 0.004 m. Look up k is you don't kow what it is. Solve for Q.
The mass won't matter in the answer to (a).
(b) This time the mass matters. Set the centripetal force m V^2/R equal to the magnetic force q V B. Thet is what keeps the particle moving in a circle.
Solve for the megnetic field B, in units of Tesla. Make sure R = 0.005 m this time.
(c)Force on a wide of length L, carrying a current I perpendicular to a field B is
F = B I L. Set this equal to the 5 N force of part (a). Solve for I
(a) To find the magnitude of each charge in part a, we can use the Coulomb force formula:
F = k * Q1 * Q2 / R^2
Where F is the electrostatic force between the charges, k is the Coulomb constant, Q1 and Q2 are the magnitudes of the charges, and R is the distance between the charges.
In this case, we have an electrostatic force of 5 N and the charges are identical, so Q1 = Q2 = Q. The distance between the charges is 4 mm, which is equal to 0.004 m.
Plugging in the values, we get:
5 = k * Q * Q / (0.004^2)
To solve for Q, we need to know the value of k. The Coulomb constant, k, is approximately 8.99 x 10^9 N m^2 / C^2.
Substituting this value into the equation, we can solve for Q:
5 = (8.99 x 10^9) * Q^2 / (0.004^2)
Q^2 = (5 * 0.004^2) / (8.99 x 10^9)
Q^2 = 2.223 x 10^-18
Taking the square root of both sides, we find:
Q = 4.712 x 10^-9 C
So, the magnitude of each charge is approximately 4.712 x 10^-9 C.
(b) To find the strength of the magnetic field in part b, we can equate the centripetal force to the magnetic force acting on the charge. The centripetal force is given by:
F_centripetal = m * V^2 / R
Where m is the mass of the charge, V is its speed, and R is the radius of the circular path.
The magnetic force acting on the charge is given by:
F_magnetic = q * V * B
Where q is the charge of the particle, V is its speed, and B is the magnetic field strength.
Setting these two forces equal to each other, we get:
m * V^2 / R = q * V * B
Plugging in the values, we have:
(0.01 * 10^-3 kg) * (50 m/s)^2 / (0.005 m) = (4.712 x 10^-9 C) * (50 m/s) * B
Simplifying the equation:
2.5 = 2.356 x 10^-7 * B
Solving for B:
B = 2.5 / (2.356 x 10^-7)
B ≈ 1.063 x 10^7 Tesla
So, the strength of the magnetic field is approximately 1.063 x 10^7 Tesla.
(c) Now, we need to find the magnitude of current that would produce the same magnetic force as the charges in part a.
The force on a wire carrying a current, length, and located in a magnetic field is given by:
F_wire = B * I * L
Where B is the magnetic field strength, I is the current, and L is the length of the wire.
We know that F_wire = 5 N, and the length of the wire is 0.5 m.
So, we have:
5 N = (1.063 x 10^7 Tesla) * I * (0.5 m)
Simplifying the equation:
10 = 5.315 x 10^6 I
Solving for I:
I = 10 / (5.315 x 10^6)
I ≈ 1.883 x 10^-6 Amps
So, the magnitude of current that would produce the same magnetic force is approximately 1.883 x 10^-6 Amps.