Question

Two rail tracks are isolated with each other and on the ground as well. They are connected with a milivoltmeter. What will be the reading in milivoltmeter when a train run at the speed of 180 km/hour on it? Given that the vertical component of earth's field is \(0.2 \times 10^{-4}\) weber/m\(^2\) and tracks are 1 m distance apart with each other.

Hint:

In motional EMF problems (\(\epsilon = Blv\)), always ensure that B, l, and v are mutually perpendicular. If they are not, you must use the appropriate components. In this problem, the train's velocity is horizontal, the axle is horizontal, and the vertical component of the magnetic field is used, fulfilling the perpendicularity condition.

Answer 1

Step 1: Understanding the Concept:
This problem describes the phenomenon of motional electromotive force (EMF). When a conductor moves through a magnetic field, an EMF is induced across its ends. Here, the axle of the train acts as the conductor, moving horizontally, and it cuts through the vertical component of the Earth's magnetic field. This induced EMF will be measured by the milivoltmeter.

Step 2: Key Formula or Approach:
The motional EMF (\(\epsilon\)) induced in a conductor of length \(l\) moving with velocity \(v\) perpendicular to a magnetic field \(B\) is given by: \[ \epsilon = B_v l v \] where \(B_v\) is the component of the magnetic field perpendicular to both the length and the velocity. In this case, it is the vertical component of the Earth's magnetic field.

Step 3: Detailed Explanation:
First, we need to ensure all units are in the SI system. \begin{itemize} \item Velocity (v): The speed of the train is given as 180 km/hour. \[ v = 180 \, \frac{\text{km}}{\text{hour}} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{hour}}{3600 \, \text{s}} = 50 \, \text{m/s} \] \item Magnetic Field (\(B_v\)): The vertical component is \(B_v = 0.2 \times 10^{-4}\) weber/m\(^2\) = \(0.2 \times 10^{-4}\) Tesla (T). \item Length (l): The distance between the tracks, which is the length of the conductor (axle), is \(l = 1\) m. \end{itemize} Now, we can calculate the induced EMF: \[ \epsilon = (0.2 \times 10^{-4} \, \text{T}) \times (1 \, \text{m}) \times (50 \, \text{m/s}) \] \[ \epsilon = (0.2 \times 50) \times 10^{-4} \, \text{V} \] \[ \epsilon = 10 \times 10^{-4} \, \text{V} = 1.0 \times 10^{-3} \, \text{V} \] Since \(1 \text{ millivolt (mV)} = 10^{-3} \text{ V}\), the induced EMF is: \[ \epsilon = 1.0 \, \text{mV} \]

Step 4: Final Answer:
The reading in the milivoltmeter will be 1.0 mV.