Question

A conductor made up of metal of length 1 m rotates vertically at angular velocity $5$ radians per second about one of its ends. Suppose the horizontal component of earth's magnetic field will be given as $0.2\times {{10}^{-4}}T$, then the e.m.f developed between the two ends of the conductor will be
$\begin{aligned} & A.5mV \\\ & B.50\mu V \\\ & C.5\mu V \\\ & D.50mV \\\ \end{aligned}$

Answer 1

The emf developed in the metal conductor will be equivalent to the half of the product of the magnetic field, angular velocity and the square of the length of the conductor. Using this equation directly will give us the answer. These will help you in solving this question.

Formula used: $e=\dfrac{B\omega {{l}^{2}}}{2}$
Where $B$ be the horizontal magnetic field component, $\omega$ be the angular velocity of its rotation and $l$ be the length of the conductor.

Complete step by step answer:
First of all let us take a look at what all are mentioned in the question. The magnetic field which is the horizontal component of earth’s magnetic field is given by the equation,
$B=0.2\times {{10}^{-4}}T$
The angular velocity with which the body is rotating can be written as,
$\omega =5\dfrac{radian}{\sec }$
And the length of the conductor has been given as,
$l=1m$
The emf produced in the conductor in between the two end of the conductor can be found by the equation,
$e=\dfrac{B\omega {{l}^{2}}}{2}$
Where $B$ be the horizontal magnetic field component, $\omega$ be the angular velocity of its rotation and $l$ be the length of the conductor.
Let us substitute the values in the equation.
$e=\dfrac{0.2\times {{10}^{-4}}\times 5\times {{1}^{2}}}{2}$
Simplifying the equation and finally we can write that,
$e=\dfrac{100\times {{10}^{-6}}}{2}=50\mu V$

So, the correct answer is “Option B”.

Note: According to Faraday's Law, we will get an induced emf if there's varying magnetic flux through a loop. If the varying emf is because of some kind motion of a conductor in a magnetic field, it can be called a motional emf.