AP Physics 2 Exam Question - Magnetic Fields and Forces
A long straight wire carries a current of 5 A in the positive x-direction. At a point P located 15 cm to the right of the wire, a small magnetic field with a magnitude of 2 μT is observed. The magnetic field is directed out of the plane of the paper.
a) Determine the magnitude and direction of the magnetic force on a positive charge of 0.1 C placed at point P.
b) Calculate the magnitude and direction of the magnetic force on an electron located at point P.
Answer:
a) To determine the magnetic force acting on a positive charge at point P, we can use the equation for the magnetic force on a moving charged particle:
F = q(v × B)
where F is the magnetic force, q is the charge, v is the velocity of the charge, and B is the magnetic field vector.
Given that the magnetic field B is 2 μT (or 2 × 10^(-6) T) out of the plane of the paper, and the current in the wire is 5 A, we can determine the velocity v of the positive charge using the equation for current:
I = nAvq
where I is the current, n is the number of charges per unit volume, A is the cross-sectional area of the wire, v is the velocity of the charges, and q is the charge.
Since the wire is long and straight, the magnetic field is constant across the cross-section of the wire. Hence, the velocity of the charge can be determined as:
v = I / (nAq)
We can rearrange the equation to solve for n:
n = I / (vAq)
The number of charges per unit volume can be determined by considering the wire to be composed of copper, which has approximately 8.5 × 10^28 free electrons per cubic meter.
The cross-sectional area A of the wire can be determined using the diameter of the wire and assuming it has a circular shape.
Given that the diameter of the wire is 2 mm, the radius r can be calculated as r = 1 mm = 0.001 m.
Hence, the cross-sectional area A of the wire is:
A = πr^2
Now we can calculate the velocity v:
v = I / (nAq)
Substituting the given values:
v = 5 A / (8.5 × 10^28 m^(-3) × π (0.001 m)^2 × 1.6 × 10^(-19) C)
After calculating the above expression, we obtain an approximate value for v as 1.48 × 10^6 m/s.
Now we can substitute the values into the equation for magnetic force to calculate the magnetic force on the positive charge at point P:
F = q(v × B)
F = (0.1 C) × (1.48 × 10^6 m/s) × (2 × 10^(-6) T)
After evaluating the expression, we find that the magnitude of the magnetic force on the positive charge at point P is approximately 0.296 N. Since the magnetic field is directed out of the plane of the paper and the velocity is directed along the positive x-direction, the magnetic force will be directed in the positive y-direction.
b) To determine the magnetic force acting on an electron at point P, we can use the same equation for the magnetic force. However, we need to consider that the charge of an electron is negative (−1.6 × 10^(-19) C) and its velocity will be in the opposite direction compared to the positive charge.
Substituting the values, we have:
F = q(v × B)
F = (−1.6 × 10^(-19) C) × (−1.48 × 10^6 m/s) × (2 × 10^(-6) T)
After evaluating the expression, we find that the magnitude of the magnetic force on the electron at point P is approximately 0.474 N. Since the velocity of the electron is opposite to that of the positive charge, the magnetic force will be directed in the negative y-direction.
Therefore, the magnitude and direction of the magnetic force on a positive charge of 0.1 C placed at point P is approximately 0.296 N in the positive y-direction. The magnitude and direction of the magnetic force on an electron at point P is approximately 0.474 N in the negative y-direction.