An equilateral triangular loop ADC moves out | vedclass

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(B) Let the side length of the equilateral triangle be $L$. As the loop moves out of the magnetic field with velocity $v$, the area $A$ of the loop in

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(B) Let the side length of the equilateral triangle be $L$. As the loop moves out of the magnetic field with velocity $v$, the area $A$ of the loop inside the magnetic field decreases.At time $t$, the height of the triangle remaining inside the magnetic field is $(h - vt)$, where $h = \frac{\sqrt{3}}{2}L$ is the total height of the triangle.The width of the triangle at a distance $x$ from the apex is $w = 2x 	an(30^{\circ}) = \frac{2x}{\sqrt{3}}$.The area $A$ inside the field at time $t$ is the area of the smaller triangle remaining inside: $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \left(\frac{2(h-vt)}{\sqrt{3}}ight) 	imes (h-vt) = \frac{(h-vt)^2}{\sqrt{3}}$.The magnetic flux is $\phi = B \cdot A = \frac{B}{\sqrt{3}}(h-vt)^2$.The induced electromotive force $(EMF)$ is $e = -\frac{d\phi}{dt} = -\frac{B}{\sqrt{3}} \cdot 2(h-vt) \cdot (-v) = \frac{2Bv}{\sqrt{3}}(h-vt)$.The induced current is $i = \frac{e}{R} = \frac{2Bv}{\sqrt{3}R}(h-vt)$.Since $i$ is a linear function of time $t$ with a negative slope, the graph of induced current versus time is a straight line with a negative slope.

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