Discussion Questions

4. When the switch is closed ===> current flows through the circuit. The two coils (solenoids) then produce a magnetic field pointed toward the left. The hammer is mounted on a magnetized bar which is then attracted to the solenoids hitting the bell. As the hammer hits the bell, it opens the circuit at the "Make-break contact" which causes the current to stop flowing ===> magnetic field generated by the solenoid goes away. The spring on the hammer then pulls it back to re-close the circuit.

Problems

36. The ball's motion and the magnetic field are initially in the same plane, the x-y plane. The magnetic field makes an angle of 23.5 degrees with respect to the initial motion of the ball (which is along the x-axis). What is v x B? Assume that "+23.5 degrees" means that the magnetic field points into the +y-direction. In this case, v x B points in the positive z-direction. (If the magnetic field pointed into the -y-direction, in which direction would v x B point?) Now, since the ball has a positive charge, the force is in the same direction as v x B.

39. The magnetic force is F = q (v x B). The proton moves perpendicularly to the field so that the magnitude of the force is F = qvB ===> B = F/(qv). Plugging in numbers, we have B = 4.0x10**(-13) N/(1.6x10**(-19) C x 2x10**6 m/s) = 1.3 T.

48. The force is F = l (I x B). The field of the Earth is horizontal but at an angle of 90 degrees with respect to the wire (current). Since the current runs west-to-east and the B-field points south-to-north, I x B ===> the force points vertically (upward). The magnitude of the force per unit length of the wire is (F/l) = IB = 10 A x 0.5 x 10**(-4) T = 5 x 10**(-4) N/m. The total force on the wire is 0.0005 N/m x 10 m = 0.005 N.

52. The torque exerted on the coil is torque = (magnetic moment) x B. The moment is (0.2 m)**2 x 1.25 A x 200 = 10 A-m**2. The field is B = 0.5 T. The maximum occurs when sine theta = 1 ===> max torque = moment x B = 10 A-m**2 x 0.5 T = 5 N-m

54. We first calculate the field due to one wire and then determine the force due to this field on the second wire.

• The field of one wire at the position of the second wire is B = (permeability/2 pi) x (I/r) = 2 x 10**(-7) x (10 A/0.25 m) = 8 x 10**(-6) T. The field is perpendicular to the wire at this point.
• The magnitude of the Force is then F = l(IB) = 50.0 m x 10 A x 8 x 10**(-6) T = 4.0 x 10**(-3) N. Since the currents are in the same direction, the force tends to pull the second wire toward the first wire (use the right hand rule to find B from the direction of I and then use the right hand rule on I x B).
• A similar argument shows that the first wire is also pulled toward the second wire (as could also be argued using Newton's third law of motion).

• Direction: the proton moves in the + x-direction. It is accelerated into the + y-direction. Because the proton has a positive charge, the force is in the direction of v x B. Using the right hand rule, the B-field must point in the - z-direction.
• Magnitude:
  • The force on the proton is F = m(proton) x acceleration = 1.7 x 10**(-27) kg x 4.0 x 10**(12) m/s**2 = 6.8 x 10**(-15) N. This force is due to the magnetic field.
  • Magnetic force = q (v x B) = 1.6x10**(-19) C x 2.0 x 10**7 m/s x B ===> B = 6.8x10**(-15) N/(3.2x10**(-12) C-m/s) = 2.1 x 10**(-3) T

65. See Figure 21.36. The magnitude of the torque is thus torque = Area x I x B x sine phi. Since the area of the coil doesn't change, and the magnitude and direction of B are constant ===> if the current is decreased by 25 %, then the torque decreases by 25 % too.

66. The coil hangs in an uniform vertical magnetic field and so the magnetic moment makes an angle of 90 degrees with the field. The torque is then torque = area x I x B. We want B ===> B = torque / (area x I) = 0.100 N-m / (pi x [0.1 m]**2 x 5.0 A) = 0.64 T.