A closed circular coil having a diameter of 50 cm made of 200 turns of wire with a total resistance of 10 Ω is placed with its plane at right angles to a magnetic field of strength of 10-2 T. Calculate the quantity of electric charge passing through it when the coil is turned through 1800 about an axis in its plane.
We know that electromotive force is
"\\displaystyle \\mathcal {E} = - \\frac{\\Delta \\Phi}{\\Delta t}"
"\\Phi = NBS" - a flux though the coil
"\\Delta \\Phi = \\Phi (cos{180^\\circ} - cos{0^\\circ})= -2 \\Phi"
By definition, current is
"\\displaystyle I = \\frac{\\Delta q}{\\Delta t}"
and according to the Ohm's law "\\displaystyle I = \\frac{\\mathcal{E}}{R}"
Combining all these formulae, we have
"\\displaystyle \\Delta q = \\frac{2 \\Phi}{R} = \\frac{2 NBS}{R} = \\frac{2 \\cdot200 \\cdot 10^{-2} \\cdot (\\pi \\cdot 0.5^2\/4)}{10} = 7.85 \\cdot 10^{-2} C"
Comments
Leave a comment