Physics 2220 Module 09 Homework

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1 Physics 2220 Module 09 Homework 01. A potential difference of V is developed across the 10-cm-long wire of the figure as it moves though a magnetic field perpendicular to the page. What are the strength and direction (in or out) of the magnetic field? Recall the equation for magnetic force on a charge: F B = q v B A positive charge is going to move to the left side of the wire due to the force acting on it as the wire moves downward through a magnetic field. By using the right hand rule, the magnetic field must be pointed out of the page. For the simple potential in the wire, the potential difference will be: Δ V = E d s = EL The potential has reached its maximum when the magnetic force is equal to the electric force. Equate these two and solve for the magnitude of the magnetic field. F B = F E qvb = qe B = Δ V vl = (0.050 V) (5.0 m/s) ( m) = 0.10 T 02. The figure shows a 2.0-cm-diameter solenoid passing through the center of a 6.0-cm-diameter loop. The magnetic field inside the solenoid is 0.20 T. What is the magnetic flux through the loop when it is perpendicular to the solenoid and when it is tilted at a 60 o angle? Note the magnetic field is uniform, but only present in the cross sectional area of the solenoid itself, not in the entire loop. Use the definition of magnetic flux. For when then loop is perpendicular: For when then loop is 60 degrees: Φ B = B d A = BA = B π r 2 cos (θ) Φ B = B π r 2 cos (θ) = (0.20 T) π ( m) 2 cos (0 o ) = Wb Φ B = B π r 2 cos (θ) = (0.20 T) π ( m) 2 cos (60 o ) = Wb

2 03. A solenoid is wound as shown in the figure. Recall that the field lines for a magnet go from north to south. (a) Is there an induced current as magnet 1 is moved away from the solenoid? If so, what is the current direction through resistor R? (1) The magnetic field in the solenoid is originally going to the left (2) The magnet moves to the right, reducing the original field (3) The loop wants to keep the original magnetic field the same and will induce a current in an attempt to do so (4) The current to produce that magnetic field will need to move from right to left across the resistor (notice what direction the solenoid is wrapped). (b) Is there an induced current as magnet 2 is moved away from the solenoid? If so, what is the current direction through resistor R? (1) The field lines will diverge in the solenoid (2) Net flux will be zero (3) The magnet moving away changes that flux but the net flux will still be zero (4) No induced current 04. The current in the solenoid of the figure is increasing. The solenoid is surrounded by a conducting loop. Is there current in the loop? If so, is the loop current cw or ccw? (1) The current is increasing, and since the magnitude of the magnetic field in the solenoid is directly proportional to the current, the magnetic field will also be increasing. (2) The induced current in the conducting loop will counter this change. (3) Current in the solenoid is counter-clockwise, so the magnetic field inside will be to the left. (4) The induced magnetic field will be to the right to counter the increasing magnetic field going to the left, so the current in the loop in the conductor will be clockwise.

3 05. The loop in the figure is being pushed into a 0.20 T magnetic field at 50.0 m/s. The resistance in the loop is 0.10 Ω. What are the direction and the magnitude of the current loop? Consider the dimensions of the loop inside the magnetic field. Use L for the length (left to right) and x for the portion of the sides that are inside the magnetic field. Notice that the magnetic fiel is uniform over all the area A of the loop in the field. Now use the Faraday's Law for this situation: ε = d Φ B The value of x is the only thing changing in time. Negative means the induced current opposed the change Use Ohm's Law to determine the current in the loop To oppose the change in the magnetic field, the magnetic field produced by the induced current will be created by a current going clockwise. 06. The resistance of the loop in the figure is 0.20 Ω. Is the magnetic field strength increasing or decreasing? At what rate (T/s)? The magnetic field is changing, but the area of the loop and the angle with the magnetic field are constant. Notice that the magnetic fiel is uniform over all the area A of the loop. Use the definition of induced EMF: = d ( B A) = B d (Lx) ε = BL dx = BLv ε = BLv = (0.20 T) ( m) (50 m/s) = 0.50 V ε = d Φ B ε = d ( B A) = A db Now use Ohm's Law to solve for the rate the magnetic field is changing. I = ε R = 0.50 V 0.10 Ω = 5.0 A db = ε A = IR A = ( A) (0.20 Ω) ( m) 2 = 4.7 T/s The direction of the induced current indicates the magnetic field created is out of the page. Since the external magnetic field is going into the page, and the induced current creates a magnetic field that goes against any change, the external magnetic field must be increasing.

4 07. The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 4.0 T/s. What is the electric field strength inside the solenoid at a point: Use Faraday's Law to derive a relationship between the changing magnetic field in the solenoid and resulting electric field. Note that the cross sectional area of the solenoid is constant and the angle between the magnetic field direction and area vector always remains at 0 degrees. Also remember the magnitude of the electric field along the line integral will be constant. (a) (b) On the axis? 2.0 cm from the axis? E = r 2 E = r 2 E d s = d Φ B E ds = A E (2 π r) = π r 2 = E = r 2 (0 m) 2 = ( m) 2 ( 4.0 T/s) = 0 V/m ( 4.0 T/s) = 0.40 V/m 08. A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60 o away from vertical increases from 0.50 T to 1.50 T in 0.60 s. What is the induced emf in the coil? Use Faraday's Law to solve for the induced EMF. A slightly modified version needs to be used to account for the fact that the coil has 100 turns. ε = N d Φ B Note that the angle between the magnetic field and the area vector is not 0 degrees and must be accounted for. Assume that the magnetic field changes uniformly over the entire area. ε = N d ( B A ) = N A cos (θ) d (B) ε = (100) (π ( m) 2 ) cos (60 o ) ( ( ) T 0.60 s ) = 26 mv

5 09. A 10-cm-diameter parallel-plate capacitor has a 1.0 mm spacing. The electric field between the plates is increasing at a rate of V / m s. What is the magnetic field strength (a) On the axis inside the capacitor Use the Ampere-Maxwell equations to solve for the magnetic field inside the capacitor: d Φ E B d s = I + ε o Note: There is no current across the capacitor (I = 0) The induced magnetic field is constant along a line tangent to the circle of radius r along the line integral The electric flux integral will be be just EA, the flux through a circle of radius r The Ampere-Maxwell equation reduces to: So on the axis: B (2 π r) = 0 + ε o A de B (2 π r) = ε o (π r 2 ) de B = ε o B = ( C 2 / N m 2 ) (4 π 10 7 ) (0 m) ( N / C s) B = 0 T r 2 de (b) (c) 3.0 cm from the axis still inside the capacitor B = ( C 2 / N m 2 ) (4 π 10 7 ) ( m) ( N / C s) B = T 7.0 cm from the axis outside the capacitor Use the Ampere-Maxwell equations to solve for the magnetic field outside the capacitor: Note: There is no current across the capacitor (I = 0) The induced magnetic field is constant along a line tangent to the circle of radius r along the line integral The electric flux integral will be be just EA, the flux through a circle of radius R (for the area of the capacitor) The Ampere-Maxwell equation reduces to: So 7.0 cm from the axis: B (2 π r) = 0 + ε o A de B (2 π r) = ε o (π R 2 ) de R 2 de B = ε o 2r B = ( C 2 / N m 2 ) (4 π 10 7 ) ( m) 2 B = T 2 ( m) ( N / C s)

6 10. A 5.0-cm-diameter parallel-plate capacitor has a 0.50 mm gap. What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V / s? The displacement current will numerically be the same as the current on the wires going to the capacitor. Reminder of the definition of capacitance, and the capacitance of a parallel-plate capacitor: The charge on the plates is changing in time: I d = dq Q = CV C C = ε o A d = d (C V C ) = C dv C I d = ( C 2 / N m 2 ) π ( m) 2 ( m) = ε o A d dv C ( V/s) = 17 μ A

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