Question

A metallic rod of length ' \(L\) ' is rotated with an angular speed of ' \(\omega\) ' normal to a uniform magnetic field ' \(B\) ' about an axis passing through one end of rod as shown in figure. The induced emf will be :

• A. \(\frac{1}{2} B ^2 L ^2 \omega\)
• B. \(\frac{1}{4} BL ^2 \omega\)
• C. \(\frac{1}{2} BL ^2 \omega\)
• D. \(\frac{1}{4} B ^2 L \omega\)

Hint:

For a rotating rod in a uniform magnetic field, the induced EMF is given by:

\[ \epsilon = \frac{1}{2} B L^2 \omega, \]

where \( B \) is the magnetic field, \( L \) is the length of the rod, and \( \omega \) is the angular speed.

Answer 1

The Correct Option is C

When the rod rotates about one end in a uniform magnetic field, the induced EMF is calculated using Faraday’s law. The differential EMF generated across an infinitesimal length \( dx \) of the rod is:

\[ d\epsilon = Bv \, dx \]

where:

• \( v = \omega x \) is the linear velocity of the element at a distance \( x \) from the axis of rotation,
• \( B \) is the uniform magnetic field.

Substitute \( v = \omega x \):

\[ d\epsilon = B(\omega x) \, dx = B\omega x \, dx \]

The total EMF is obtained by integrating \( d\epsilon \) along the length of the rod:

\[ \epsilon = \int_0^L B\omega x \, dx \]

\[ \epsilon = B\omega \int_0^L x \, dx \]

\[ \epsilon = B\omega \left[ \frac{x^2}{2} \right]_0^L = B\omega \frac{L^2}{2} \]

Thus, the total induced EMF is:

\[ \epsilon = \frac{1}{2} BL^2 \omega \]

Answer 2

The correct answer is (C) : $\frac{1}{2} BL ^2 \omega $

∫dε=∫B(ωx)dx 
ε=Bω0∫L​xdx=2BωL2​