Final answer:
The emf induced between the two sides of a car traveling at 34 m/s through the Earth's magnetic field with a vertically downward component of 5.3×10−5 T across a width of 2.1 m is 3.7494 mV, with positive charge accumulating on the driver's side.
Step-by-step explanation:
To find the emf induced between the two sides of a car traveling through the Earth's magnetic field, we can use Faraday's Law of Induction, which states that the emf (ε) induced in a circuit is proportional to the rate of change of the magnetic flux through the circuit. In this scenario, the car itself acts as a moving conductor through the Earth's magnetic field and thus a voltage is generated across the width of the car as a result of its motion through the field. To calculate this, the formula ε = vBL is used, where 'v' is the velocity of the car, 'B' is the magnetic field strength, and 'L' is the width of the car:
ε = (34 m/s) × (5.3×10−5 T) × (2.1 m) = 3.7494×10−3 V or 3.7494 mV.
Since the positive side of the voltage is where the conventional current exits and we're considering a conventional current (positive charge flow), and if the car is moving perpendicularly to the magnetic field, we'd expect the positive charge to accumulate on the driver's side of the car if we assume the car is moving forward in such a way that the right-hand rule indicates positive on the left. Therefore, the emf induced is a positive 3.7494 mV.