Question

A milli voltmeter is connected in parallel to an axle of the train running with a speed of$180\dfrac{{km}}{{hr}}$. If the vertical component of earth’s magnetic field is $0 \cdot 2 \times {10^{ - 4}}\dfrac{{Wb}}{{{m^2}}}$ and the distance between the rails is 1m, then the reading of voltmeter will be:
A) ${10^{ - 2}}volt$.
B) ${10^{ - 4}}volt$.
C) ${10^{ - 3}}volt$.
D) $1volt$.

Answer 1

Magnetic field is defined as the magnetic influence on a moving charge, electric currents and magnetic materials. Voltmeter is a device which is used to measure the potential difference between two points.

Formula used: The formula used for measuring the voltage is given by,
$V = B \cdot l \cdot v$
Where voltage is V the magnetic field is B the length is l and velocity is v.

Complete step by step answer:
It is given that a milli voltmeter is connected in parallel to an axle of the train running with a speed of$180\dfrac{{km}}{{hr}}$ the vertical component of earth’s magnetic field is $0 \cdot 2 \times {10^{ - 4}}\dfrac{{Wb}}{{{m^2}}}$ and the distance between the rails is 1m we need to find the reading of voltmeter.
As the magnetic field is equal to$0 \cdot 2 \times {10^{ - 4}}\dfrac{{Wb}}{{{m^2}}}$, the speed of the train is $180\dfrac{{km}}{{hr}}$ and length is 1m.
Also let us convert the speed into$\dfrac{m}{s}$,
$\Rightarrow v = \left( {180 \times \dfrac{5}{{18}}} \right)\dfrac{m}{s}$
$\Rightarrow v = 50\dfrac{m}{s}$………eq. (1)
The formula used for measuring the voltage is given by,
$V = B \cdot l \cdot v$
Where voltage is V the magnetic field is B the length is l and velocity is v.
$\Rightarrow V = B \cdot l \cdot v$………eq. (2)
Replace the values of the magnetic field, the length. Replace the value of velocity of the axle from equation (2)
$\Rightarrow V = \left( {0 \cdot 2 \times {{10}^{ - 4}}} \right) \times \left( 1 \right) \times \left( {50} \right)$
$\Rightarrow V = {10^{ - 3}}volt$.
The reading of the voltmeter is equal to$V = {10^{ - 3}}volt$.

The correct answer for this problem is option C.

Note: It is advisable for students to remember the formula of the voltage developed for a moving coil as it can be helpful for solving these kinds of problems. The units should be similar while solving for the voltage and due to the magnetic field and moving coil. For converting the unit from $km hr{.^{ - 1}}$ to $m{s^{ - 1}}$ just multiply the velocity with$\dfrac{5}{{18}}$.