Question

A long, thin solenoid has 950 turns per meter and radius 2.00 cm . The current in the solenoid is increasing at a uniform rate of 61.0 A/s .

Part A

What is the magnitude of the induced electric field at a point 0.460 cm from the axis of the solenoid?

E1 =

___

  V/m  

Part B

What is the magnitude of the induced electric field at a point 1.20 cm from the axis of the solenoid?

E2 =

___

  V/m  

Answer 1

Part-A

From Faradey's law,

now, Magnetic field inside a solenoid is given by,

So,

given, r = 0.460 cm = 0.460*10^-2 m

n = 950 turns/m

= 4*pi*10^-7

dI/dt = +61.0 A/s

So, E = induced emf = 0.5*(0.460*10^-2)*(4*pi*10^-7)*950*61.0

E = 1.67*10^-4 V/m

Part-B

now, r = 1.20 cm

then, Magnetic field inside a solenoid(because point is inside the solenoid) is given by,

So,

E = induced emf = 0.5*(1.20*10^-2)*(4*pi*10^-7)*950*61.0

E = 4.37*10^-4 V/m

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