Question

A uniform magnetic field exists in the region given by B$ = 3\hat{i} + 4\hat{j} + 5\hat{k}$. A rod of length 5 m placed along the y-axis is moved along the x-axis with constant speed 1m/sec. Then induced emf in the rod will be
(a) zero
(b) 25 V
(c)20 V
(d) 15 V

Answer 1

Hint: To find out the induced emf on a conductor, the component of magnetic field(B) which is perpendicular to length (l) and velocity (v) is taken into account (here the z component). Induced emf is given as $e = - \dfrac{{d\phi }}{{dt}}$. Where, $\phi = B.A$.

Step by step answer: Given the magnetic field B$ = 3\hat{i} + 4\hat{j} + 5\hat{k}$
The length of the rod, l$ = 5m$ and its velocity, v$ = 1m/s$.
We know that according to Faraday's law the induced EMF is negative of the rate of change of flux. $e = - \dfrac{{d\phi }}{{dt}}$. Where, $\phi$ is the flux $ = B.A$
The negative sign is neglected as it shows the direction which is opposite to the direction of change of flux.
$e = \dfrac{{d\left( {BA} \right)}}{{dx}}$
As the magnetic field B is constant and area ( A) is equal to the product of length (l) and breadth (b) of the rod, we get the induced EMF as $e = B\dfrac{{d\left( {lb} \right)}}{{dt}}$$ = Bl\dfrac{{db}}{{dt}}$
We know that it is the velocity of the rod v$ = \dfrac{{db}}{{dt}}$.
Substituting the value, we get, $e = Blv$.
In this equation we assume that B, l and v are mutually perpendicular to each other and therefore the component of magnetic field (B) which is perpendicular to length (l) and velocity (v) is taken into account.
The length is along the y axis and the velocity is along the x axis therefore the magnetic field component of the z-axis is used.
So, $e = 5 \times 5 \times 1 = 25V$

Hence induced emf e=25V correct option is B.

Note: To find out the induced EMF on a current carrying conductor the direction of its length element, its velocity and the magnetic field should be mutually perpendicular to each other. We can find the direction of induced EMF using lenz law.