Question

A 1.2 m wide railway track is parallel to the magnetic meridian. The vertical component of the earth’s magnetic field is 0.5 Gauss. when a train runs on the rails at a speed of 60 km/hr, then the induced potential difference the ends of the axle will be
A. \[{10^{ - 4}}\,{\text{V}}\]
B. \[2 \times {10^{ - 4}}\,{\text{V}}\]
C. \[{10^{ - 3}}\,{\text{V}}\]
D. Zero

Answer 1

Hint: Use Faraday’s law of magnetic induction. \[\varepsilon = N\dfrac{{d\phi }}{{dt}}\]
Here, N is the number of turns and \[\phi \] is the magnetic flux. And, \[\phi = BA\cos \theta \]
Here, B is the magnetic field, A is the area vector and \[\theta \] is the angle between the magnetic field and area vector.

Complete step by step answer:
To find the magnitude of induced emf or induced potential difference along the axle, we can use Faraday’s law of induction as follows,
\[\varepsilon = N\dfrac{{d\phi }}{{dt}}\]

Here, N is the number of turns and \[\phi \] is the magnetic flux.

In this problem, \[N = 1\].

Therefore,
\[\varepsilon = \dfrac{{d\phi }}{{dt}}\] …… (1)

Also, the total magnetic flux is expressed as,
\[\phi = BA\cos \theta \]
Here, B is the magnetic field, A is the area vector and \[\theta \] is the angle between the magnetic field and area vector.
In this case, since the magnetic field of the earth is vertical and the railway track is horizontal. Therefore, the angle between the magnetic field and area vector is \[90^\circ \]. Therefore, substitute \[90^\circ \] for \[\theta \] in the above equation.
\[\phi = BA\cos 90^\circ \]
\[ \Rightarrow \phi = BA\left( 1 \right)\]

Substitute \[BA\] for \[\phi \] in equation (1).
 \[\varepsilon = \dfrac{{d\left( {BA} \right)}}{{dt}}\]
\[ \Rightarrow \varepsilon = B\dfrac{{dA}}{{dt}}\] …… (2)

The area swept by the axle of the train is
\[\Delta A = lx\]

Here, l is the width of the railway track and x is the distance covered by the railway.
Substitute \[lx\]for \[\Delta A\]in equation (2).
 \[\varepsilon = B\dfrac{{d\left( {lx} \right)}}{{dt}}\]
                                                                                                                                                  …… (3)
But we know that the velocity is equal to rate of change of distance with time, therefore,
\[v = \dfrac{{dx}}{{dt}}\]
Therefore, equation (3) becomes,
\[\varepsilon = Blv\]
Substitute \[0.5\,{\text{Gauss}}\] for B, 1.2 m for l and \[60\,{\text{km/hr}}\] for v in the above equation.
\[\varepsilon = \left( {\left( {0.5\,{\text{Gauss}}} \right)\left( {\dfrac{{1\,{\text{T}}}}{{{{10}^4}\,{\text{Gauss}}}}} \right)} \right)\left( {1.2\,{\text{m}}} \right)\left( {\left( {60\,{\text{km/hr}}} \right)\left( {\dfrac{{1\,{\text{m}}}}{{{{10}^{ - 3}}\,{\text{km}}}}} \right)\left( {\dfrac{{1\,{\text{hr}}}}{{3600\,{\text{s}}}}} \right)} \right)\]
\[\therefore \varepsilon = {10^{ - 3}}\,{\text{V}}\]

So, the correct answer is Option C.

Note:
The S.I unit of magnetic field is Tesla and the vertical component of earth’s magnetic field is given in Gauss. Always convert the unit of magnetic field into Tesla and other quantities into S.I. units to use the formula \[\varepsilon = Blv\].