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A metallic rod of length 'I' is tied to a string of length 2I and made to rotate with angular speed $\omega$ on a horizontal table with one end of the string fixed.If there is a vertical magnetic field 'B' in the region,the e.m.f. induced across the ends of the rod is :

Option: 1
$\frac{5B\omega I^{2}}{2}$

Option: 2
$\frac{2B\omega I^{2}}{2}$

Option: 3
$\frac{3B\omega I^{2}}{2}$

Option: 4
$\frac{4B\omega I^{2}}{2}$

Answers (1)

best_answer

Motional E.m.f due to rotational motion -

Conducting rod $\rightarrow$

$\varepsilon =\frac{1}{2}Bl^{2}\omega =Bl^{2}\pi\nu$

$\nu \rightarrow f\! r\! equency$

$T \rightarrow Time\; period$

- wherein

Consider a small length ax on the rod. emf across this lenth

d $\varepsilon = B .\upsilon. dx$

= $B \omega x dx.$

total emf = $\int_{2l}^{3l} B\omega x dx$

= $\frac{B\omega }{2} : \left [ \left ( 3l \right )^{2}-\left ( 2l \right )^{2} \right ]$

= $\frac{5}{2} B \omega l^{2}$

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