Answer the following questions based on the diagram below: 16. Choose the correct statements (1) loop ABCD will rotate along pqr (2) loop ABCD will rotate along rqp (3) loop ABCD will oscillate back and forth (4) split ring will rotate along with the loop (A) 1, 3 (B) 3, 4 (C) 2, 3 (D) 1, 4 17. The diagram is a representation of a simple (A) dc motor (B) ac motor (C) dc generator (D) ac generator 18. Choose the correct statements (1)The direction of current through the loop reverses after every half-rotation (2)The direction of current through the loop does not change (3)We must use Fleming’s Right Hand rule to predict the direction of force on segments AB and CD (4)No magnetic force acts on segment BC when the coil is in the position shown in the diagram (A) 2, 3 (B) 2, 4 (C) 1, 3 (D) 1, 4 For heater rated at 4kW and 220 V, calculate 19. the current (A) 10 A (B) 18.18 A (C) 4.4 (D) 28 A 20. The resistance of the heater (A) 12.1 Ω (B) 6.05 Ω (C) 24.2 Ω (D) 36.3 Ω 21. The energy consumed in 2 hours (A) 4 kWh (B) 16 kWh (C) 2 kWh (D) 8 kWh
Answers
Answer:
Torque on a Current Loop: Motors and Meters
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Describe how motors and meters work in terms of torque on a current loop.
Calculate the torque on a current-carrying loop in a magnetic field.
Motors are the most common application of magnetic force on current-carrying wires. Motors have loops of wire in a magnetic field. When current is passed through the loops, the magnetic field exerts torque on the loops, which rotates a shaft. Electrical energy is converted to mechanical work in the process. (See Figure 1.)
Diagram showing a current-carrying loop of width w and length l between the north and south poles of a magnet. The north pole is to the left and the south pole is to the right of the loop. The magnetic field B runs from the north pole across the loop to the south pole. The loop is shown at an instant, while rotating clockwise. The current runs up the left side of the loop, across the top, and down the right side. There is a force F oriented into the page on the left side of the loop and a force F oriented out of the page on the right side of the loop. The torque on the loop is clockwise as viewed from above.
Figure 1. Torque on a current loop. A current-carrying loop of wire attached to a vertically rotating shaft feels magnetic forces that produce a clockwise torque as viewed from above.
Let us examine the force on each segment of the loop in Figure 1 to find the torques produced about the axis of the vertical shaft. (This will lead to a useful equation for the torque on the loop.) We take the magnetic field to be uniform over the rectangular loop, which has width w and height l. First, we note that the forces on the top and bottom segments are vertical and, therefore, parallel to the shaft, producing no torque. Those vertical forces are equal in magnitude and opposite in direction, so that they also produce no net force on the loop. Figure 2 shows views of the loop from above. Torque is defined as τ = rF sin θ, where F is the force, r is the distance from the pivot that the force is applied, and θ is the angle between r and F. As seen in Figure 2(a), right hand rule 1 gives the forces on the sides to be equal in magnitude and opposite in direction, so that the net force is again zero. However, each force produces a clockwise torque. Since r = w/2, the torque on each vertical segment is (w/2)F sin θ, and the two add to give a total torque.
Diagram showing a current-carrying loop from the top, and four different times as it rotates in a magnetic field. The magnetic field oriented toward the right, perpendicular to the vertical dimension of the loop. In figure a, the top view of the loop is oriented at an angle to the magnetic field lines, which run left to right. The force on the loop is up on the lower left side where the current comes out of the page. The force is down on the upper right side where the loop goes into the page. The angle between the force and the loop is theta. Torque is clockwise and equals w over 2 times I l B sine theta. Figure b shows the top view of the loop parallel to the magnetic field lines. The force on the loop is up on the left side where I comes out of the page. The force on the loop is down on the right side where I goes into the page. The angle theta between the F and B is ninety degrees. Torque is clockwise and equals w over 2 I l B equals maximum torque. Figure c shows the top view of the loop oriented perpendicular to B. The force on the loop is up at the top, where I comes out of the page, and down at the bottom where I goes into the page. Theta equals 0 degrees. Torque equals zero since sine theta equals 0. In figure d the force is down on the lower left side of the loop where I goes in, and up on the upper right side of the loop where I comes out. The torque is counterclockwise. Torque is negative.
Figure 2.
CONCEPTUAL QUESTIONS
1. Draw a diagram and use RHR-1 to show that the forces on the top and bottom segments of the motor’s current loop in Figure 1 are vertical and produce no torque about the axis of rotation.
PROBLEMS & EXERCISES
1. (a) By how many percent is the torque of a motor decreased if its permanent magnets lose 5.0% of their strength? (b) How many percent would the current need to be increased to return the torque to original values?
2. (a) What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.0-A current in a 1.60-T field? (b) What is the torque when θ is 10.9º?
3. Find the current through a loop needed to create a maximum torque of 9.00 N. The loop has 50 square turns that are 15.0 cm on a side and is in a uniform 0.800-T magnetic field.
4. Calculate the magnetic field strength needed on a 200-turn square loop 20.0 cm on a side to create a maximum torque of 300 N ⋅ m if the loop is carrying 25.0 A.
5. Since the equation for torque on a current-carrying loop is
, the units of N ⋅ m must equal units of A ⋅ m2 T. Verify this.