A square loop of side \(10\mathit{cm}\) and resistance \(0.5\Omega \) is placed vertically in the east-west plane. A uniform magnetic field of \(0.1T\) is setup across the plane in the north-east direction. The magnetic field is decreased to zero in \(0.70s\) at a steady rate. Determine the magnitudes of induced emf and current during this time-interval.
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A square loop of side \(10\mathit{cm}\) and resistance \(0.5\Omega \) is placed vertically in the east-west plane. A uniform magnetic field of \(0.1T\) is setup across the plane in the north-east direction. The magnetic field is decreased to zero in \(0.70s\) at a steady rate. Determine the magnitudes of induced emf and current during this time-interval.
Answer
1mV, 2mA
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Solution
To solve this question, we need to calculate the initial flux before the magnetic field becomes zero and final flux after the field gets zero. According to question, initially the field makes an angle of \(45{}^{\circ}\) with the area vector of the square loop.
Area of the square loop \(\left(A\right)=(0.1m)\)
The initial flux,
\(\therefore \;{\varphi }_{i}={B}{A}\cos 45{}^{\circ }\)
\(=0.1\times 10^{-2}\times \frac 1{\sqrt 2}\)
\(=0.70\times 10^{-3}\) Wb
The final flux, \(\varphi _f=0\)
Therefore, the magnitude of the emf induced is given by electromagnetic induction;
\(\therefore \;\varepsilon =\left\vert \frac{\Delta \varphi }{\Delta t}\right\vert\)
\(=\left|\frac{\varphi _f-\varphi _i}{{\Delta}t}\right|\)
\(=\left|\frac{0-0.70\times 10^{-3}}{0.70}\right|\)
\(=1\times 10^{-3}V=1\mathit{mV}\)
And, the magnitude of the current is
\(\therefore \;i=\frac{\varepsilon }{R}=\frac{{10}^{−3}}{0\mathrm{\ldotp }5}=2{m}{A}\)
Area of the square loop \(\left(A\right)=(0.1m)\)
The initial flux,
\(\therefore \;{\varphi }_{i}={B}{A}\cos 45{}^{\circ }\)
\(=0.1\times 10^{-2}\times \frac 1{\sqrt 2}\)
\(=0.70\times 10^{-3}\) Wb
The final flux, \(\varphi _f=0\)
Therefore, the magnitude of the emf induced is given by electromagnetic induction;
\(\therefore \;\varepsilon =\left\vert \frac{\Delta \varphi }{\Delta t}\right\vert\)
\(=\left|\frac{\varphi _f-\varphi _i}{{\Delta}t}\right|\)
\(=\left|\frac{0-0.70\times 10^{-3}}{0.70}\right|\)
\(=1\times 10^{-3}V=1\mathit{mV}\)
And, the magnitude of the current is
\(\therefore \;i=\frac{\varepsilon }{R}=\frac{{10}^{−3}}{0\mathrm{\ldotp }5}=2{m}{A}\)
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