INTRO: The effects due to the interaction of a current-carrying loop with a magnetic field have many applications, some as common as the electric motor. This problem illustrates the basic principles of this interaction.
Figure 1
Figure 2
Figure 3
Figure 4
Consider a current I that flows in a plane rectangular current loop with height a = 4.00 cm and horizontal sides b = 2.00 cm . (Figure 1) The loop is placed into a uniform magnetic field B⃗ in such a way that the sides of length a are perpendicular to B⃗ , and there is an angle θ between the sides of length b and B⃗ , as shown in the figures. (Figure 2) PART A:
PART B:
For parts B and C, the loop is initially positioned at θ=30°.
Assume that the current flowing into the loop is 0.500A . If the magnitude of the magnetic field is 0.300T , what is τ, the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field?
Express your answer in newton-meters.
First off we have to determine the magnitude of the force on each vertical bar, which they will be inversely equivalent. So, using the formula F = ILB, F = (0.5 A)(0.04 m)(0.3 T) = 6×10-3 N NOW, since the ring is skewed, we have to disregard the component of the force that is along the 'lever'. Using our handy right triangle, we can determine that the perpendicular component of the force will be F1 = F cos(θ) = 5.2×10-3 N Now, recalling from the beginning of physics, torque = F×l τ1 = (5.2×10-3 N) ⋅ (0.01 m) [we get this l from half the length of the top bar] τ1 = 5.2×10-5 N⋅m but this isn't our answer quite yet, remember they asked for the net torque. What we just found was the torque for the 'upper' half of the ring. To get our net torque, we must add the torque from the 'top' to the torque from the 'bottom' luckily, the two torques are equal so we just multiply our previous value by 2 and determine that the net torque is 1.04×10-4 N⋅m
PART C:
What happens to the loop when it reaches the position for which θ=90∘, that is, when its horizontal sides of length b are perpendicular to B⃗ (see the figure)? (Figure 3)
The direction of rotation changes because the net torque acting on the loop causes the loop to rotate in a clockwise direction.
B)
C)
D)
so once the ring rotates to the 90° point, there will be no more force on the edges of it, however, there won't be any forces on the other side either, resulting in it to continue rotating until something stops it, which is satisfied by B).
PART D:
Now suppose that you change the initial angular position of the loop relative to B⃗ , and assume that the loop is placed in such a way that initially the angle between the sides of length b and B⃗ is θ=120, as shown in the figure. (Figure 4) Will the interaction of the current through the loop with the magnetic field cause the loop to rotate?
B)
C)
D)
Same idea as in the first problem, except in the opposite direction. This means that there will be torque on the ring just in the other way now. So, A).
The figure shows a wire that is connected to a power supply and suspended between the poles of a magnet. When the switch is closed, the wire deflects in the direction shown.
Which of the dashed boxes A–D represents the position of the north magnetic pole?
SOLUTION: As he explains in the video, if you point your right thumb towards yourself to represent the direction of the current, and then your pointer finger down to represent the magnetic field pointing from north to south, the middle finger will represent the direction that the wire is displaced. Working backward, pointing your right thumb toward yourself and middle finger down in the direction of deflection, your pointer finger points to the left. Since this represents the magnetic field from north to south, the dashed box that represents the north magnetic pole is therefore C.
INTRO: You are given two infinite, parallel wires each carrying current I. The wires are separated by a distance d, and the current in the two wires is flowing in the same direction. This problem concerns the force per unit length between the wires.
PART A: Is the force between the wires attractive or repulsive?
SOLUTION: The second right hand rule assists us in solving this problem. Since we have two parallel wires with equal current going the same direction, that vector represents each of the wires velocity vectors. By pointing our right thumb in this direction and using our pointer finger to represent the magnetic field (pointing back toward ourselves), at any point, the Force of the wire will be represented by our middle finger. Thinking of this for both wires, the force vectors are pointing towards each other, resulting in an attractive relationship.
PART B: What is the force per unit length F/L between the two wires? Express your answer in terms of I, d, and constants such as μ0 and π.
SOLUTION: First some relevant formulas: FB=qvBsin(θ) = qv→×B→ on a long straight wire, F=ILBsin(θ) = I→L×B→ where L is the length of the wire So the magnetic field at wire 2 from the current in wire 1 will be B=μ0I1 ⋅ 1/[2πd] The force on a length ΔL of wire 2 will be F=ΔLI2×B The force per unit length in terms of the currents will be, F/ΔL = μ0I1I2 ⋅ 1/[2πd] since I1 = I2, F/L = μ0I2 / [2πd]
PART C: In the SI system, the unit of current, the ampere, is defined by this relationship using an apparatus called an Ampère balance. What would be the force per unit length of two infinitely long wires, separated by a distance 1m, if 1A of current were flowing through each of them? Express your answer numerically in newtons per meter.
SOLUTION: If I = 1 A, & d = 1 m, plugging these values into the formula from the previous section, F/L = μ0(1 A)2 / [2π(1m)] = μ0/ [2π] ⋅ A2/m μ0 = 4π×10-7 ⋅ T⋅m/A → 4π×10-7 / [2π] ⋅A2/m ⋅T⋅m/A → 2×10-7 ⋅ T⋅A 1 T = 1 N / [A⋅m] →2×10-7 ⋅ N/m
