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Standard XII

Physics

Coupling

An equilatera...

Question

An equilateral triangular loop having a resistance R and length of each side l is placed in a magnetic field which is varying at dB/dt = 2ms^-1The induced current in the loop is,

$\frac{\sqrt{3} I^{2}}{R}$

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$\frac{\sqrt{3} I^{2}}{2 R}$

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$\frac{\sqrt{3} I^{2}}{2 R}$

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$\frac{\sqrt{3} I^{2}}{5 R}$

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Solution

The correct option is B
$\frac{\sqrt{3} I^{2}}{2 R}$

$Magnetic flux, φ = \frac{\sqrt{3} l^{2} B}{4} But, E = \frac{- d ϕ}{d t}, | E | = \frac{\sqrt{3} l^{2} B}{4} \frac{d B}{d t} = \frac{\sqrt{3} l^{2} (2)}{4} \Rightarrow i = \frac{E}{R} = \frac{\sqrt{3} I^{2}}{2 R}$

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An equilateral triangular loop having a resistance R and length of each side l is placed in a magnetic field which is varying at dB/dt = 2ms-1 The induced current in the loop is,

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An equilateral triangular loop having a resistance R and length of each side l is placed in a magnetic field which is varying at dB/dt = 2ms^-1The induced current in the loop is,

The correct option is B
$\frac{\sqrt{3} I^{2}}{2 R}$

$Magnetic flux, φ = \frac{\sqrt{3} l^{2} B}{4} But, E = \frac{- d ϕ}{d t}, | E | = \frac{\sqrt{3} l^{2} B}{4} \frac{d B}{d t} = \frac{\sqrt{3} l^{2} (2)}{4} \Rightarrow i = \frac{E}{R} = \frac{\sqrt{3} I^{2}}{2 R}$