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A Barlow's wheel is a simple motor similar to Faraday's disk. A source of EMF Ue = 3.00 V supplies current through sliding contact K1 (at axis) and mercury contact K2 (at disk rim). Rheostat resistance ranges from R1 = 3.00 ohm to R2 = 6.00 ohm. Magnetic field B = 800 mT covers the disk area. Disk: radius r0 = 120 mm, moment of inertia J = 5.00 x 10⁻⁴ kg m². Mechanical resistance ignored. Which of the following statements are correct?

1. The direction of rotation of the wheel when viewed from the source toward the north pole is anti-clockwise.
2. The maximum angular velocity of the wheel is independent of the rheostat resistance.
3. The initial angular acceleration for R = R1 = 3 ohm is 11.52 rad/s².
4. The initial angular acceleration for R = R2 = 6 ohm is 5.76 rad/s².

Correct answer: The initial angular acceleration for R = R1 = 3 ohm is 11.52 rad/s².

Solution

For a conducting disk in field B, torque = (1/2)*B*I*r0². At t=0 (omega=0), back EMF=0. I = Ue/R. For R=R1=3: I=3/3=1 A. tau = 0.5*0.8*1*(0.12)² = 0.5*0.8*0.0144 = 0.00576 N*m. alpha = tau/J = 0.00576/(5e-4) = 11.52 rad/s². Option C is correct. For R=R2=6: I=3/6=0.5 A. tau=0.5*0.8*0.5*0.0144=0.00288 N*m. alpha=0.00288/5e-4=5.76 rad/s². Option D is also correct. Max angular velocity: at max omega, net torque=0. Back-EMF of disk = (1/2)*B*omega*r0². Voltage equation: Ue = I*R + (1/2)*B*omega*r0². At max speed torque=0 means current*force=0... actually at steady state for an ideal motor without friction: Ue = back-EMF = (1/2)*B*omega_max*r0². So omega_max = 2*Ue/(B*r0²) = 2*3/(0.8*0.0144) = 6/0.01152 = 520.8 rad/s. This is independent of R! So option B is correct. Options C and D are both numerically correct.

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