Conducting rods moving in a magnetic field and a capacitor circuit

Explanation

Induced Electromotive Force and Circuit Equation

When a conducting rod is moving to the right at a speed $v$v, the magnitude of the induced electromotive force $V$V generated in the conducting rod by electromagnetic induction is expressed as follows: $V = vBL$V=vBL According to Lenz's law and the right-hand rule, this induced electromotive force is generated in a direction that tends to cause current to flow counterclockwise through the circuit. If the current flowing through the circuit is $I$I and the charge stored in the capacitor is $q$q, then Kirchhoff's second law (circuit equation) holds as follows:

$vBL - \frac{q}{C} - RI = 0$vBL−Cq​−RI=0

Equation of Motion and Time Progression

When the current $I$I flows through the conducting rod, the conducting rod is subjected to an Ampere force of magnitude $F_A = IBL$FA​=IBL directed to the left (opposing the motion) from the magnetic field. Therefore, the equation of motion for the conducting rod is as follows:

$m \frac{dv}{dt} = F - IBL$mdtdv​=F−IBL

Here, we differentiate both sides of the circuit equation with respect to time $t$t. Using the relationship between current and charge $I = \frac{dq}{dt}$I=dtdq​, we obtain the following equation:

$BL \frac{dv}{dt} - \frac{I}{C} - R \frac{dI}{dt} = 0$BLdtdv​−CI​−RdtdI​=0

After a sufficient amount of time has passed, the magnitude of the current approaches a constant value $I$I, so the rate of change of the current over time becomes negligible ($\frac{dI}{dt} \to 0$dtdI​→0). Also, at this time, the acceleration of the conducting rod $\frac{dv}{dt}$dtdv​ approaches a constant value $a$a.

Therefore, the differentiated circuit equation above settles into the following relationship:

$BLa - \frac{I}{C} = 0 \implies a = \frac{I}{CBL}$BLa−CI​=0⟹a=CBLI​

Calculation of Current

The equation of motion in this case is expressed as $ma = F - IBL$ma=F−IBL, so we substitute the calculated acceleration $a$a.

$m \left( \frac{I}{CBL} \right) = F - IBL$m(CBLI​)=F−IBL

Multiply both sides by $CBL$CBL and rearrange for $I$I.

$mI = FCBL - ICB^2L^2$mI=FCBL−ICB2L2

$I(m + CB^2L^2) = FCBL$I(m+CB2L2)=FCBL

$I = \frac{FCBL}{m + CB^2L^2}$I=m+CB2L2FCBL​

Numerical Substitution

We substitute the given numerical constraints and perform the calculation.

• Denominator term: $CB^2L^2 = 0.500 \times 2.00^2 \times 0.400^2 = 0.500 \times 4.00 \times 0.160 = 0.320 \text{ kg}$CB2L2=0.500×2.002×0.4002=0.500×4.00×0.160=0.320 kg
• Entire denominator: $m + CB^2L^2 = 0.200 + 0.320 = 0.520 \text{ kg}$m+CB2L2=0.200+0.320=0.520 kg
• Numerator term: $FCBL = 3.00 \times 0.500 \times 2.00 \times 0.400 = 1.20 \text{ A}\cdot\text{ kg}$FCBL=3.00×0.500×2.00×0.400=1.20 A⋅ kg

Therefore, the current $I$I is as follows:

$I = \frac{1.20}{0.520} = \frac{120}{52} = \frac{30}{13} \text{ A}$I=0.5201.20​=52120​=1330​ A

Comparing this to the irreducible fraction $\frac{p}{q}$qp​, we get $p = 30$p=30 and $q = 13$q=13.

The value we are looking for is $p + q = 30 + 13 = 43$p+q=30+13=43.

Answer: 43