Part of a single rectangular loop of wire with dimensions .35 m by .75 m (.35 is the height and .75 is the length) is situated inside a region of uniform magnetic field of 0.395 T. The total resistance of the loop is 0.690 Ω.
Calculate the force required to pull the loop from the field (to the right) at a constant velocity of 6.36 m/s. Neglect gravity.
Calculate the rate of magnetic flux decrease when withdrawing the loop from the B-field at that velocity. The field will be out of the loop when it has moved 0.75 m, which will take t = 0.118 s if the loop is moved along the length dimension. Calculate the energy disipated in the resistor during that time. It is
(V^2/R)*t
The voltage is
V = A*B/t, where A is the loop area.
The energy disipated in the resistor while moving the loop equals the force times the distance it moves. Solve for the force.
The induced emf in the loop is
ε =dΦ/dt =d(BwL)/dt=Bwv,
The current is
I = ε/R = Bwv/R
The force is
F =(Bw)^2v/R
(w =0.35 m, B = 0.395 T, v =6.36 m/s,
R= 0.690 Ω)
To calculate the force required to pull the loop from the magnetic field, we can use the following equation:
F = B * I * L
Where:
- F is the force required
- B is the magnetic field strength
- I is the current flowing through the loop
- L is the length of the wire in the magnetic field
Now, we need to find the current flowing through the loop. We can use Ohm's Law to calculate it:
V = I * R
Where:
- V is the voltage across the loop
- I is the current flowing through the loop
- R is the resistance of the loop
In this case, the voltage across the loop is equal to the velocity of the loop multiplied by the height of the loop. So, we have:
V = v * h
Where:
- v is the velocity of the loop
- h is the height of the loop
Now, substitute this equation into Ohm's Law equation:
v * h = I * R
Solving for I, we get:
I = (v * h) / R
Substituting this value of I in the force equation, we have:
F = B * (v * h) / R * L
Now, you can substitute the values given in the question to calculate the force required.