A 40.0-turn coil of wire of radius 3.0 cm is placed between the poles of electro
ID: 2009823 • Letter: A
Question
A 40.0-turn coil of wire of radius 3.0 cm is placed between the poles of electromagnet which produces magnetic field directed to the north. The field increases from 0 to 0.75 T at a constant rate in a time interval of 225 s. What is the magnitude and direction of induced current in the coil if the coil has 1O resistance and the field is perpendicular to the plane of the coil? Make a sketch showing direction of magnetic field, orientation of coil and direction of flow of current. Do not need to draw all 40 turns, just one turn will work.Explanation / Answer
Given: Radious of the coil wire = r = 3 cm = 0.03 m Area of the coil = A = r2 = (3.14)(0.03)2 A = 0.002826 m2 Rate of change of Magnetic field with time as : d B / dt = (0.75 T - 0T )/ (225 s) = 0.0033 T / s Resistence of the coil = R = 10 ohm No. of turns of the ciol is here taken as N = 1 not 40 It is known by the formula induced emf : emf = - N A dB/dt = - (1)(0.002826 )(0.0033) = 0.0000093258 volt (magnitude) thus, current : I = e / R = 0.00000093258 A or 0.93258A Direction: The direction induced current will be opposite to the change of magnetic field direction acc. to Lenz's law Resistence of the coil = R = 10 ohm No. of turns of the ciol is here taken as N = 1 not 40 It is known by the formula induced emf : emf = - N A dB/dt = - (1)(0.002826 )(0.0033) = 0.0000093258 volt (magnitude) thus, current : I = e / R = 0.00000093258 A or 0.93258A Direction: The direction induced current will be opposite to the change of magnetic field direction acc. to Lenz's lawRelated Questions
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