A metallic rod of length l is tied to a string of length 2l and made to rotate with angular speed ω on a horizontal table with one end of the string fixed. If there is a vertical magnetic field B in the region, the

Correct answer: B. 5Bω l ^ 2 2 e = ∫ _ 2l ^ 3l BVdx e = ∫ _ 2 ℓ ^ 3 ℓ B(ωx)dx e = ∫ _ 2l ^ 3l xdx e = [ x ^ 2 2 ] _ 2l ^ 3l e = ( )/(2) [ (3l) ^ 2 - (2l) ^ 2 ] _ 2l ^ 3l e = (5)/(2)ω Bl ^ 2

JEE Main 20137 April

Exam: JEE Main
Year: 2013
Date: 7 April
Subject: Physics
Chapter: Electromagnetic Induction

JEE Main 2013 · 7 April · Physics · Electromagnetic Induction

A metallic rod of length l is tied to a string of length 2l and...

A metallic rod of length $l$ is tied to a string of length $2l$ 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:

A metallic rod of length l is tied to a string of length 2l and made to rotate with angular speed ω on a horizontal

Options

A $\dfrac{4\text{B}\omega{l}^{2}}{2}$
B $\dfrac{5\text{B}\omega{l}^{2}}{2}$
C $\dfrac{2\text{B}\omega{l}^{2}}{2}$
D $\dfrac{3\text{B}\omega{l}^{2}}{2}$

Correct answer

B. $\dfrac{5\text{B}\omega{l}^{2}}{2}$

Step-by-step solution

A metallic rod of length l is tied to a string of length 2l and made to rotate with angular speed ω on a horizontal

$e = \mathop{∫}_{2l}^{3l}BVdx$

$\text{e} = \mathop{∫}_{2 ℓ}^{3 ℓ}$ B(ωx)dx
$e = \omegaB\mathop{∫}_{2l}^{3l}xdx$

$e = \omegaB{\left[\dfrac{{x}^{2}}{2}\right]}_{2l}^{3l}$

$e = \dfrac{\omegaB}{2}{\left[{\left(3l\right)}^{2} - {\left(2l\right)}^{2}\right]}_{2l}^{3l}$

$e = \dfrac{5}{2}\omega{Bl}^{2}$

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