GSO002 Problem 15
Problem Statement
Magnetic Energy Transition in Electromagnetic Induction Coupled with LC Resonance
Problem Statement
On a horizontal surface, two sufficiently long parallel metal rails are fixed at separation d. Let the near rail be P and the far rail be Q.
One plate of a capacitor with capacitance C is connected to the left end of rail P; the other plate is connected via a switch and a coil of self-inductance L in series to the left end of rail Q.
A metal bar of mass m is placed across the two rails perpendicularly. The bar can slide smoothly without friction while maintaining electrical contact with the rails.
A uniform downward magnetic field of flux density B is applied to the entire system. The electrical resistance of the bar, rails, wire, and coil are all negligible.
Initially, the bar is at rest and the switch is open. The capacitor is charged to Q0 with the plate connected to rail P being positive.
At time t=0, the switch is gently closed. The discharging capacitor exerts an Ampere force on the bar, and the bar starts to move on the rails.
During the subsequent motion, the magnetic energy stored in the coil fluctuates over time.
Let Umax be the maximum magnetic energy stored in the coil during the motion. Find the value of Umax expressed in millijoules (mJ).
Constraints
- Mass of bar: m=0.070 kg
- Rail separation: d=0.50 m
- Magnetic flux density: B=2.0 T
- Capacitance: C=0.040 F
- Self-inductance: L=0.15 H
- Initial charge: Q0=0.22 C
Input Format
Give the value of Umax as a positive integer.
Solution
1. Basic Relation Between Current and Bar Velocity
Let q be the charge on the P-side plate at time t (initially q(0)=Q0>0). Let i=−dtdq be the circuit current (positive = clockwise). The bar experiences an Ampere force leftward, so with leftward positive:
mdtdv=idB=−dtdqdB2. Momentum-Charge Conservation
Integrating from t=0 to t (with initial velocity v(0)=0):
mv=dB(Q0−q)⋯(1)3. Energy Conservation
Total energy is conserved (no resistance):
2CQ02=2Cq2+UL+21mv2Substituting v=mdB(Q0−q) from equation (1):
UL=2CQ02−2Cq2−2m(dB)2(Q0−q)24. Maximizing UL
Define f(q)=2Cq2+2m(dB)2(Q0−q)2. Minimizing f(q):
f′(q)=Cq−m(dB)2(Q0−q)=0 Q0−q=m+C(dB)2mQ0Substituting back:
fmin=E0⋅m+C(dB)2C(dB)2Therefore:
Umax=E0−fmin=E0⋅m+C(dB)2m(Note: The self-inductance L does not appear in the maximum value expression — it only affects the time scale.)
5. Numerical Calculation
E0=2CQ02=2×0.040(0.22)2=0.0800.0484=0.605 J dB=0.50×2.0=1.0 T⋅m C(dB)2=0.040×(1.0)2=0.040 kg m+C(dB)2m=0.1100.070=117 Umax=0.605×117=0.385 J=385 mJAnswer: 385