A solenoid (coil) having 200 turns, and length 10cm. It is carrying a current of

Question

A solenoid (coil) having 200 turns, and length 10cm. It is carrying a current of 5A. It has a cross section area of the core of 6x10^-5 m^2. 1). What is the inductance of the coil? 2). What is the magnetic field at the center of the coil? 3). What is the magnetic flux? 4). What is the energy stored in the solenoid? 5). If the current in the solenoid is changing at the rate of 50 A/s, what is the induced EMF in the coil? A solenoid (coil) having 200 turns, and length 10cm. It is carrying a current of 5A. It has a cross section area of the core of 6x10^-5 m^2. 1). What is the inductance of the coil? 2). What is the magnetic field at the center of the coil? 3). What is the magnetic flux? 4). What is the energy stored in the solenoid? 5). If the current in the solenoid is changing at the rate of 50 A/s, what is the induced EMF in the coil? 1). What is the inductance of the coil? 2). What is the magnetic field at the center of the coil? 3). What is the magnetic flux? 4). What is the energy stored in the solenoid? 5). If the current in the solenoid is changing at the rate of 50 A/s, what is the induced EMF in the coil?

Explanation / Answer

N=200, L=0.1m, I= 5A, A=6*10^-5 m^2, dI/dt=50A/s

1)

Inductance L= (µ0N2A)/L = (4*10^-7*200^2*6*10^-5)/0.1 = 3.02*10^-5 H

2)

Magnetic field B = (µ0NI)/L = (4*10^-7*200*5)/0.1 = 0.0126 T

3)

Magnetic flux = BA = 0.0126*6*10^-5 = 7.56*10^-7 Tm^2

4)

Energy stored U = ½*LI^2 = ½*3.02*10^-5*5^2 = 0.00038 J

5)

EMF = d/dt = d(BA)/dt = A*dB/dt = A*d[(µ0NI)/L]/dt = [(µ0N)/L]*dI/dt = [(4*10^-7*200)/0.1]*50 = 0.126 V

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