A metallic rod of length L is rotated with an angular speed of ω...

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 metallic rod of length L is rotated with an angular speed of ω normal to a uniform magnetic field B about an axis .

Options

A $\dfrac{1}{2}{B}^{2}{L}^{2}\omega$
B $\dfrac{1}{4}B{L}^{2}\omega$
C $\dfrac{1}{2}B{L}^{2}\omega$
D $\dfrac{1}{4}{B}^{2}L\omega$

Correct answer

C. $\dfrac{1}{2}B{L}^{2}\omega$

Step-by-step solution

A metallic rod of length L is rotated with an angular speed of ω normal to a uniform magnetic field B about an axis

Linear velocity at the distance $x$ from the centre can be written as, $v = \omegax$ .

Therefore, induced EMF due to small element will be,

$dε = Bvdx = B\left(\omegax\right)dx$

Integrating both sides, we get

$∫ dε = ∫ B\left(\omegax\right)dx$

$⇒ ε = B\omega{∫}_{0}^{L}xdx = \dfrac{B\omega{L}^{2}}{2}$

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