A conducting bar of length L rotates in a counterclockwise direction with a constant angular speed of +2.0 rad/s about a pivot P at one end, as shown. A uniform magnetic field (magnitude = 0.20 T) is directed into the paper. If L = 0.40 m, what is the potential difference, VA - VB?
To find the potential difference, VA - VB, we need to consider the motion of the conducting bar in the magnetic field. Based on the given information, we can solve the problem by following these steps:
Step 1: Determine the induced emf in the conducting bar.
According to Faraday's law of electromagnetic induction, the induced emf (ε) in a conducting bar moving in a magnetic field is given by the equation:
ε = -BLv, where B is the magnetic field strength, L is the length of the bar, and v is the velocity of the bar.
Substituting the given values:
ε = -0.20 T * 0.40 m * 2.0 rad/s
= -0.16 V
Step 2: Determine the potential difference, VA - VB.
The potential difference, VA - VB, is the same as the induced emf (ε) because the conducting bar is a closed circuit. So:
VA - VB = ε
= -0.16 V
Therefore, the potential difference, VA - VB, is -0.16 V.
To determine the potential difference between points A and B, we can use the equation for the magnetic force on a moving charge in a magnetic field:
F = q * v * B * sin(θ)
Where:
F is the magnetic force
q is the charge of the particle
v is its velocity
B is the magnetic field strength
θ is the angle between the velocity and the magnetic field direction.
In this case, the moving charges are the electrons in the conducting bar, and the force due to the magnetic field will cause an accumulation of charge at the ends of the bar, creating a potential difference.
To find the potential difference, we need to find the velocity of the electrons at points A and B and then use the equation mentioned earlier.
Given that the conducting bar is rotating in a counterclockwise direction with a constant angular speed of +2.0 rad/s, the velocity of the electrons at point A and B is different.
At point A: The velocity of the electrons is perpendicular to the magnetic field since point A is at the bottom of the rotating bar.
At point B: The velocity of the electrons is along the length of the bar and parallel to the magnetic field since point B is at the top of the rotating bar.
Let's calculate the velocities:
At point A:
The distance from the pivot P to point A is L/2, so the tangential velocity (v_A) at point A is given by:
v_A = ω * (L/2)
= 2.0 rad/s * (0.40/2) m
= 0.80 m/s
At point B:
The tangential velocity (v_B) at point B is given by:
v_B = ω * R
= 2.0 rad/s * 0.40 m
= 0.80 m/s
Now, we can calculate the potential difference using the equation:
VA - VB = (q * vA * B * sin(θ)) / q - (q * vB * B * sin(θ)) / q
Since we only need the potential difference, we can ignore the charge (q) and the angle (θ) since they cancel out.
VA - VB = vA * B - vB * B
= B * (vA - vB)
= 0.20 T * (0.80 m/s - 0.80 m/s)
= 0 T
Therefore, the potential difference VA - VB is 0 volts.