Question

A metal rod of length $1\,{\text{m}}$ is rotated about one of its ends in a plane at right angle to a uniform magnetic field of induction $2.5 \times {10^{ - 3}}\,W{\text{b}} \cdot {{\text{m}}^{ - 2}}$. If it rotates at $1800\,{\text{rpm}}$, then the induced emf between its ends approximately is:
A. ${\text{0}}{\text{.24 V}}$
B. ${\text{0}}{\text{.12 V}}$
C. ${\text{0}}{\text{.36 V}}$
D. ${\text{0}}{\text{.48 V}}$

Answer 1

Use the formula for the emf induced by the rotating object in the magnetic field and the formula for the linear speed of the object in the circular motion. Then derive an equation for the emf induced in terms of the magnetic field, length of the object and the angular speed of the object.

Formulae used:

The induced emf $e$ of an object moving in a magnetic field is
$e = BLv$ …… (1)

Here, $B$ is the magnetic field, $L$ is the length of the moving object and $v$ is the speed of the moving object.

The linear velocity of an object in circular motion is
$v = R\omega$ …… (2)

Here, $R$ is the radius of the circular path and $\omega$ is the angular speed of the object.

Complete step by step answer:
The metal rod of length $1\,{\text{m}}$ is rotated about one of its ends in a plane at right angle to a uniform magnetic field of induction $2.5 \times {10^{ - 3}}\,W{\text{b}} \cdot {{\text{m}}^{ - 2}}$.

The rod is rotating with the angular speed of $1800\,{\text{rpm}}$.

Calculate the emf induced between the ends of the metal rod.

Convert the unit of the angular speed in the SI system of units.
$\omega = \left( {1800\,{\text{rpm}}} \right)\left( {\dfrac{{2\pi }}{{60}}} \right)\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}$
$\Rightarrow \omega = \left( {1800\,{\text{rpm}}} \right)\left( {\dfrac{{2 \times 3.14}}{{60}}} \right)\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}$
$\Rightarrow \omega = 188.4\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}$

Hence, the angular speed of the rotating end of the rod is $188.4\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}$.

Rewrite the equation (2) for the linear speed $v$ of the one end of the rod moving around its other end.
$v = \dfrac{L}{2}\omega$

Here, $L$ is the length of the rod which is the diameter for the circular path for the rotating end of the rod and $\omega$ is the angular speed of the rod end.

Substitute $\dfrac{L}{2}\omega$ for $v$ in equation (1).
$e = BL\left( {\dfrac{L}{2}\omega } \right)$
$\Rightarrow e = \dfrac{{B{L^2}\omega }}{2}$

Substitute $2.5 \times {10^{ - 3}}\,W{\text{b}} \cdot {{\text{m}}^{ - 2}}$ for $B$, $1\,{\text{m}}$ for $L$ and $188.4\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}$ for $\omega$ in the above equation.
$\Rightarrow e = \dfrac{{\left( {2.5 \times {{10}^{ - 3}}\,W{\text{b}} \cdot {{\text{m}}^{ - 2}}} \right){{\left( {1\,{\text{m}}} \right)}^2}\left( {188.4\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}} \right)}}{2}$
$\Rightarrow e = 235.5 \times {10^{ - 3}}\,{\text{V}}$
$\Rightarrow e = 0.235\,{\text{V}}$
$\Rightarrow e \approx 0.24\,{\text{V}}$

Therefore, the voltage induced between the ends of the rod is $0.24\,{\text{V}}$

Hence, the correct option is A.

Note: In the present problem, one end of the rod is moving around its other end. Hence, the radius of the circular path of the rotating end is half of the length of the rod.