Problem#1
The plane of a 5.0 cm x 8.0 cm rectangular loop of wire is parallel to a 0.19-T magnetic field. The loop carries a current of 6.2 A. (a) What torque acts on the loop? (b) What is the magnetic moment of the loop? (c) What is the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field?Answer:
Known:
Magnetic field, B = 0.19 T
Current, I = 6.2 A
Area, A = 5.0 cm x 8.0 cm = 0.05 m x 0.08 m
(a) torque acts on the loop is
t = IBA sin φ
t = IBwh sin φ
t = (6.2 A)(0.19 T)(0.05 m x 0.08 m) sin 900 = 4.7 x 10-3 N.m
(b) the magnetic moment of the loop is
μ = IA = Iwh = 6.2 A x 0.05 m x 0.08 m = 0.025 A.m2
(c) the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field is when maximum area is when the loop is circular with radius
R = (0.050 m + 0.080 m)/π = 0.0414 m
A = πR2 = π(0.0414 m)2 = 5.38 x 10-3 m2 and
τ = IBA = (6.2 A)(0.19 T)(5.38 x 10-3 m2) = 6.34 x 10-3 N.m
Problem#2
The 20.0 cm x 35.0 cm rectangular circuit shown in Fig. 1 is hinged along side ab. It carries a clockwise 5.00-A current and is located in a uniform 1.20-T magnetic field oriented perpendicular to two of its sides, as shown. (a) Draw a clear diagram showing the direction of the force that the magnetic field exerts on each segment of the circuit (ab, bc, etc.). (b) Of the four forces you drew in part (a), decide which ones exert a torque about the hinge ab. Then calculate only those forces that exert this torque. (c) Use your results from part (b) to calculate the torque that the magnetic field exerts on the circuit about the hinge axis ab.
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| Fig.1 |
Known:
Magnetic field, B = 1.20 T
Current, I = 5.0 A
Area, A = 20.0 cm x 35.0 cm = 0.2 m x 0.35 m
(a) diagram showing the direction of the force that the magnetic field exerts on each segment of the circuit (ab, bc, etc.) shown by the Figure 2.
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| Fig.2 |
(b) the force that produces torque for ab hinges is the force on the cd wire namely Fcd and the magnitude is
Fcd = ILB sin θ = 5.0 A x 0.20 m x 1.20 T sin 900 = 1.20 N
(c) the torque that the magnetic field exerts on the circuit about the hinge axis ab given by
τ = Fcd x Lda sin 900 = 1.20 N x 0.35 m = 0.420 N.m
Problem#3
A rectangular coil of wire, 22.0 cm by 35.0 cm and carrying a current of 1.40 A, is oriented with the plane of its loop perpendicular to a uniform 1.50-T magnetic field, as shown in Fig. 3. (a) Calculate the net force and torque that the magnetic field exerts on the coil. (b) The coil is rotated through a 30.00 angle about the axis shown, with the left side coming out of the plane of the figure and the right side going into the plane. Calculate the net force and torque that the magnetic field now exerts on the coil. (Hint: In order to help visualize this three-dimensional problem, make a careful drawing of the coil as viewed along the rotation axis.)
(c) the torque that the magnetic field exerts on the circuit about the hinge axis ab given by
τ = Fcd x Lda sin 900 = 1.20 N x 0.35 m = 0.420 N.m
Problem#3
A rectangular coil of wire, 22.0 cm by 35.0 cm and carrying a current of 1.40 A, is oriented with the plane of its loop perpendicular to a uniform 1.50-T magnetic field, as shown in Fig. 3. (a) Calculate the net force and torque that the magnetic field exerts on the coil. (b) The coil is rotated through a 30.00 angle about the axis shown, with the left side coming out of the plane of the figure and the right side going into the plane. Calculate the net force and torque that the magnetic field now exerts on the coil. (Hint: In order to help visualize this three-dimensional problem, make a careful drawing of the coil as viewed along the rotation axis.)
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| Fig.3 |
Answer:
Current, I = 1.40 A
magnetic field, B = 1.50 T
Area, A = 22.0 cm x 35.0 cm = 0.22 m x 0.35 m
(a) the net force and torque that the magnetic field exerts on the coil is F1 + F2 = 0 and F3 + F4 = 0, the forces on each side of the coil are shown in Figure 4a. The net force on the coil is zaro. Because, φ = 0, and sin φ = 0, so τ = 0. The forces on the coil produce no torque.
(b) The net force is still zero. When φ = 300 and the net torque is
t = IBA sin φ
τ = 1.40 A x 0.220 m x 0.350 m x 1.50 T x sin 30.00 = 0.0808 N.m
The net torque is clockwise in Figure 4b and is directed so as to increase the angle φ.
magnetic field, B = 1.50 T
Area, A = 22.0 cm x 35.0 cm = 0.22 m x 0.35 m
(a) the net force and torque that the magnetic field exerts on the coil is F1 + F2 = 0 and F3 + F4 = 0, the forces on each side of the coil are shown in Figure 4a. The net force on the coil is zaro. Because, φ = 0, and sin φ = 0, so τ = 0. The forces on the coil produce no torque.
(b) The net force is still zero. When φ = 300 and the net torque is
t = IBA sin φ
τ = 1.40 A x 0.220 m x 0.350 m x 1.50 T x sin 30.00 = 0.0808 N.m
The net torque is clockwise in Figure 4b and is directed so as to increase the angle φ.
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| Fig.4 |
Problem#4
A uniform rectangular coil of total mass 212 g and dimensions 0.500 m x 1.00 m is oriented with its plane parallel to a uniform 3.00-T magnetic field (Fig. 5). A current of 2.00 A is suddenly started in the coil. (a) About which axis (A1 or A2) will the coil begin to rotate? Why? (b) Find the initial angular acceleration of the coil just after the current is started.
A uniform rectangular coil of total mass 212 g and dimensions 0.500 m x 1.00 m is oriented with its plane parallel to a uniform 3.00-T magnetic field (Fig. 5). A current of 2.00 A is suddenly started in the coil. (a) About which axis (A1 or A2) will the coil begin to rotate? Why? (b) Find the initial angular acceleration of the coil just after the current is started.
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| Fig.5 |
Answer:
Known:
Magnetic field, B = 3.00 T
Current, I = 2.0 A
Area, A = 0.500 m x 1.00 m
Mass, m = 212 g = 0.212 kg
(a) About which axis (A1 or A2) will the coil begin to rotate? Why?
Since t = IBA sin f then only the top and bottom sides will experience torque, thus will rotate about A2.
(b) the initial angular acceleration of the coil just after the current is started is
Using Newton’s Second law,
∑τ = Iα
Now, each 1.00 m side has a mass of 0.0707 kg. And each 0.500-m side has a mass of 0.0353 kg.
To find the moment of inertia of the coil, consider each side as a uniform bar. The 2 0.5-m sides rotate about their centers, and the 2 1.0-m sides rotate about a parallel axis. Thus,
I = 2 x (0.0707 kg)(0.250 m)2 + 2 x (1/12)(0.0353 kg)(0.500 m)2 = 0.01031 kg.m2
│τ│ = IAB sin φ = (2.0 A)(0.500 m x 1.00 m)(3.00 T) sin 900 = 3.00 N.m
Then,
∑τ = Iα
α = ∑τ/I = 3.00 N.m/0.01031 kg.m2 = 290.98 rad/s2
Problem#5
A circular coil with area A and N turns is free to rotate about a diameter that coincides with the x-axis. Current I is circulating in the coil. There is a uniform magnetic field B in the positive y-direction. Calculate the magnitude and direction of the torque τ and the value of the potential energy U as given in U = –μ x B = –μB cos φ, when the coil is oriented as shown in parts (a) through (d) of Fig. 6.
Magnetic field, B = 3.00 T
Current, I = 2.0 A
Area, A = 0.500 m x 1.00 m
Mass, m = 212 g = 0.212 kg
(a) About which axis (A1 or A2) will the coil begin to rotate? Why?
Since t = IBA sin f then only the top and bottom sides will experience torque, thus will rotate about A2.
(b) the initial angular acceleration of the coil just after the current is started is
Using Newton’s Second law,
∑τ = Iα
Now, each 1.00 m side has a mass of 0.0707 kg. And each 0.500-m side has a mass of 0.0353 kg.
To find the moment of inertia of the coil, consider each side as a uniform bar. The 2 0.5-m sides rotate about their centers, and the 2 1.0-m sides rotate about a parallel axis. Thus,
I = 2 x (0.0707 kg)(0.250 m)2 + 2 x (1/12)(0.0353 kg)(0.500 m)2 = 0.01031 kg.m2
│τ│ = IAB sin φ = (2.0 A)(0.500 m x 1.00 m)(3.00 T) sin 900 = 3.00 N.m
Then,
∑τ = Iα
α = ∑τ/I = 3.00 N.m/0.01031 kg.m2 = 290.98 rad/s2
Problem#5
A circular coil with area A and N turns is free to rotate about a diameter that coincides with the x-axis. Current I is circulating in the coil. There is a uniform magnetic field B in the positive y-direction. Calculate the magnitude and direction of the torque τ and the value of the potential energy U as given in U = –μ x B = –μB cos φ, when the coil is oriented as shown in parts (a) through (d) of Fig. 6.
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| Fig.6 |
Answer:
The magnitude of τ is equal to the magnitude of μ x B or τ = μ x B, and we can conclude immediately that the corresponding potential energy is U = –μB cos φ, where μ = NIB, τ = μB sin φ.
(a) For Figure 5a, φ = 900, τ = NIAB sin 900 = NIAB, direction k x j = –i, and U = –μB cos 900 = 0
(b) For Figure 5b, φ = 00, τ = NIAB sin 00 = 0, no direction and U = –μB cos 00 = –NIAB
(c) For Figure 5c, φ = 900, τ = NIAB sin 900 = NIAB, direction –k x j = +i, and U = –μB cos 900 = 0
(c) For Figure 5d, φ = 1800, τ = NIAB sin 1800 = 0, no direction and U = –μB cos 1800 = NIAB
Problem#6
A coil with magnetic moment 1.45 A.m2 is oriented initially with its magnetic moment antiparallel to a uniform 0.835 T magnetic field. What is the change in potential energy of the coil when it is rotated 1800 so that its magnetic moment is parallel to the field?
Answer:
Known:
Magnetic moment, μ = 1.45 A.m2
magnetic field, B = 0.835 T
the change in potential energy of the coil when it is rotated 1800 so that its magnetic moment is parallel to the field given by
ΔU = U2 – U1
ΔU = μB Δcosφ
ΔU = (1.45 A.m2)(0.835 T)(cos 1800 – cos 00)
ΔU = –2.425 J
(a) For Figure 5a, φ = 900, τ = NIAB sin 900 = NIAB, direction k x j = –i, and U = –μB cos 900 = 0
(b) For Figure 5b, φ = 00, τ = NIAB sin 00 = 0, no direction and U = –μB cos 00 = –NIAB
(c) For Figure 5c, φ = 900, τ = NIAB sin 900 = NIAB, direction –k x j = +i, and U = –μB cos 900 = 0
(c) For Figure 5d, φ = 1800, τ = NIAB sin 1800 = 0, no direction and U = –μB cos 1800 = NIAB
Problem#6
A coil with magnetic moment 1.45 A.m2 is oriented initially with its magnetic moment antiparallel to a uniform 0.835 T magnetic field. What is the change in potential energy of the coil when it is rotated 1800 so that its magnetic moment is parallel to the field?
Answer:
Known:
Magnetic moment, μ = 1.45 A.m2
magnetic field, B = 0.835 T
the change in potential energy of the coil when it is rotated 1800 so that its magnetic moment is parallel to the field given by
ΔU = U2 – U1
ΔU = μB Δcosφ
ΔU = (1.45 A.m2)(0.835 T)(cos 1800 – cos 00)
ΔU = –2.425 J
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