We can solve this problem using the concept of electromagnetic induction and the Lorentz force law. Here’s a step-by-step breakdown:

1. Induced EMF (Electromotive Force):

Since the rod is sliding to the right in a magnetic field, it induces an EMF due to the motion of the rod in the presence of the magnetic field (Faraday's law of induction).

The induced EMF $\mathcal{E}$ is given by:

$\mathcal{E} = B \cdot L \cdot v$

where:

• $B = 0.694 \, \text{T}$ (magnetic field strength),
• $L = 0.379 \, \text{m}$ (distance between the rails),
• $v$ is the velocity of the rod (which we need to find later).

2. Current in the Circuit:

The current $I$ in the circuit due to the induced EMF is related by Ohm’s Law:

$I = \frac{\mathcal{E}}{R}$

where:

• $I = 0.148 \, \text{A}$ (current in the resistor),
• $R = 12.5 \, \Omega$ (resistance of the resistor).

Substituting the expression for $\mathcal{E}$ into Ohm’s law:

$0.148 = \frac{B \cdot L \cdot v}{12.5}$

3. Solve for Velocity $v$:

Rearranging the equation to solve for $v$:

$v = \frac{0.148 \cdot 12.5}{B \cdot L}$

Substituting the given values:

$v = \frac{0.148 \cdot 12.5}{0.694 \cdot 0.379}$

$v \approx 6.93 \, \text{m/s}$

4. Force to Maintain Constant Velocity:

To maintain the rod’s constant velocity, a force must be applied to counteract the magnetic force (Lorentz force) acting on the rod due to the current.

The magnetic force $F$ on the rod is given by:

$F = I \cdot L \cdot B$

Substituting the values:

$F = 0.148 \cdot 0.379 \cdot 0.694$

$F \approx 0.0389 \, \text{N}$

Final Answer:

The force that must be exerted on the rod to maintain a constant current of 0.148 A is approximately 0.0389 N.

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Would you like more details or have any further questions?

5 Related Questions:

1. How would the required force change if the magnetic field were doubled?
2. What would happen if the resistor value were reduced?
3. How does the velocity affect the induced EMF in this setup?
4. What if the magnetic field were pointing into the page instead of out of the page?
5. Can we calculate the power dissipated in the resistor in this scenario?

Tip: Always apply Faraday’s law to link motion in magnetic fields to induced currents in conductors!