Question

a rod of length 2l is rotating about one of its ends with angular velocity ω. If a magnetic field B₀ exists in parallel with axis of rotation, then find emf developed across the end points of rod.
(A)1/2 B₀l^2ω
(B)B₀l^2ω
(C)3/2 B₀l^2ω
(D)2 B₀l^2ω

Answer 1

Concept Introduction:-

It could take the shape of a word or a numerical representation of a quantity's arithmetic value.

Given Information:-

We have been given that a rod of length $2l$ is rotating about one of its ends with angular velocity ω. If a magnetic field $B_0$ exists in parallel with axis of rotation

To Find:-

We have to find that emf developed across the end points of rod

Solution:-

According to the problem

Consider an infinitesimally small length of rod at a distance $\mathrm{x}$ from the center of the circular path.

The emf induced across this element $\mathrm{d} \mathcal{E}=\mathrm{B_0v}(\mathrm{dx})$ $\mathrm{d} \mathcal{E}=\mathrm{B_0}(\mathrm{x} \omega)(\mathrm{dx})=\mathrm{B_0wx} \mathrm{x}$

Thus the net EMF induced $\mathcal{E}=\int_{0}^{1} \mathrm{~B_0} \omega \mathrm{xdx}$

$\begin{aligned}&\therefore \mathcal{E}=\mathrm{B_0w} \int_{0}^{1} \mathrm{xdx}=\mathrm{B_0w} \times\left.\frac{\mathrm{x}^{2}}{2}\right|_{0} ^{1} \\&\Rightarrow \mathcal{E}=\mathrm{B_0w} \times \frac{\left(1^{2}-0\right)}{2}=\frac{\mathrm{B_02l}^{2} \omega}{2}=B_0l^2\omega\end{aligned}$

Final Answer:-

The correct option is $B_0l^2\omega$.

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