Week #8 Problems

  1. A square loop 24.0 cm on a side has a resistance of 6.50 Ω. It is initially on a 0.755-T magnetic field,, with its plane perpendicular to B, but is removed from the field in 40.0 ms. Calculate the electric energy dissipated in this process.
    Solution.
    The average induced emf is
    E = -ΔФB/Δt = -AΔB/Δt = -(0.240 m)2(0 - 0.755 T)/(0.0400 s) = 1.09 V
    The average current is
    I =E /R = (1.09 V)/(6.50 Ω) = 0.168 A.
    The energy dissipated is
    E = I2RΔt = (1.68 A)2(6.50 Ω)(0.0400 s) = 7.33 x 10-3 J.
  2. Two conducting rails 30 cm apart rest on a 5.0° ramp. They are joined at the bottom by a 0.60-Ω resistor, and at the top a copper bar of mass 0.040 kg is laid across the rails. The whole apparatus is immersed in a vertical 0.55 T field. What is the terminal (steady) velocity of the bar as it slides frictionlessly down the rails?
    Solution.
    The component of the velocity of the bar that is perpendicular to the magnetic field is vcos θ, so the induced emf is 
          E  = BLvcos θ
    This produces a current in the wire
           I = E /R = (BLvcos θ)/R into the page.
    Because the current is perpendicular to the magnetic field, the force on the wire from the magnetic field will be horizontal as shown, with magnitude
           FB = ILB = (B2L2vcos θ)/R
    For the wire to slide down at a steady speed, the net force must be zero. If we consider  the components along the rail, we have
           FBcos θ - mgsin θ = 0, or
           [(B2L2vcos θ)/R]cos θ = B2Lvcos2 θ)/R = mgsin θ
           (0.55 T)2(0.30 m)2v(cos2 5°)/(0.60 Ω) = 0.040 kg)(9.8 m/s2)sin5°,
     which gives v = 0.76 m/s.    		  
  3. High-intensity desk lamps are rated at 40 W but require only 12 V. They contain a transformer that converts 120 V household voltage. (a) Is the transformer step-up or step-down? (b) what is the current in the secondary when the lamp is on? (c) What is the current in the primary? (d) What is the resistance of the bulb when on?
    Solution.
    (a) Because Vs < Vp, this is a step-down transformer.
    (b) We assume 100% efficiency, so we find the current in the secondary from
            P = IsVs;
            40 W = Is(12 V), which gives Is = 3.3 A.
    (c) We find the current in the primary from
           P = IpVp;
           40 W = Ip(120 V), which gives Ip = 0.33 A.
    (d) We find the resistance of the bulb from
           Vs = IsRs;
          120 V = (3.33 A)Rs, which gives Rs = 3.6 Ω.