Question

An axle or truck is $2.5m$ long. If the truck is moving due North at $30m/s$ at a place where the vertical component of the earth's magnetic field is $90\mu T$, the potential difference between the two ends of the axle is

Answer 1

Step 1: Given data

1. The length of the track is $L=2.5m.$
2. The velocity of the track is $V=30m/s.$
3. The vertical component of the magnetic field of the earth $B=90\times {10}^{-6}T.$

Step 2: Formula used

1. We know, that the induced emf in a conductor of length $L$ and moving with a velocity $V$ in a magnetic field $B$ is defined by the form, $e=BVL$.
2. In an open circuit, the induced emf is equal to the potential difference.

Step 3: Find the potential difference

Let the potential difference between the two ends of the truck is $e$,

Now, the potential difference is

$$\begin{gathered} e=BVL=\left(90\times {10}^{-6}\right)\times 30\times 2.5 \\ ore=6.75\times {10}^{-3}volt. \end{gathered}$$

Therefore, the potential difference between the two ends of the axle is $e=6.75\times {10}^{-3}volt.$