An equilateral triangular loop of side L is carrying current i. Suddenly a magnetic field of magnitude B is switched on parallel to one of the sides of loop. Maximum angular velocity of loop if its mass is m is

\sqrt{\frac{6 i B}{m}}

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2 \sqrt{\frac{i B}{m}}

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\sqrt{\frac{3 \sqrt{3} i B}{m}}

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\sqrt{\frac{6 \sqrt{3} i B}{m}}

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Solution

The correct option is D $\sqrt{\frac{6 \sqrt{3} i B}{m}}$

By conservation of energy
$T .1 ε_{i} = T .1 ε_{F} \Rightarrow 0 = - μ B + \frac{1}{2} I ω^{2}$
$i \frac{\sqrt{3}}{4} L^{2} B = \frac{1}{2} \frac{m L^{2}}{12} ω^{2}$
$ω = \sqrt{\frac{6 \sqrt{3} i B}{m}}$

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An equilateral triangular loop of side L is carrying current i. Suddenly a magnetic field of magnitude B is switched on parallel to one of the sides of loop. Maximum angular velocity of loop if its mass is m is

The correct option is D √6√3 i Bm By conservation of energy T.1εi=T.1εF⇒0=−μB+12Iω2 i√34L2B=12mL212ω2 ω=√6√3 i Bm

The correct option is D $\sqrt{\frac{6 \sqrt{3} i B}{m}}$

By conservation of energy
$T .1 ε_{i} = T .1 ε_{F} \Rightarrow 0 = - μ B + \frac{1}{2} I ω^{2}$
$i \frac{\sqrt{3}}{4} L^{2} B = \frac{1}{2} \frac{m L^{2}}{12} ω^{2}$
$ω = \sqrt{\frac{6 \sqrt{3} i B}{m}}$