Complete Question
The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 4.60 T/s.
Part A What is the electric field strength inside the solenoid at a point on the axis?
Part B
What is the electric field strength inside the solenoid at a point 1.50 cm from the axis?
Answer:
Part A
[tex]E = 0 \ V/m[/tex]
Part B
[tex]E_{15} = 0.0345 \ V/m[/tex]
Explanation:
From the question we are told that
The diameter of the solenoid is [tex]d = 5.0 \ cm = 0.05 \ m[/tex]
The magnetic field is [tex]B = 2.0 \ T[/tex]
The rate of the change of the magnetic field is [tex]\frac{dB}{dt} = 4.60 \ T/s[/tex]
The radius of the solenoid is mathematically represented as
[tex]R = \frac{ d}{2}[/tex]
substituting values
[tex]R = \frac{ 5.0 *10^{-2}}{2} = 0.025 \ m[/tex]
Generally the of the solenoid is mathematically represented as
[tex]E = \frac{ r}{2} * |\frac{dB}{dt} |[/tex]
Now at the point on axis is r = 0 given that the axis is the origin so
[tex]E = \frac{ 0}{2} * |\frac{dB}{dt} |[/tex]
[tex]E = 0 \ V/m[/tex]
Now the electric field strength inside the solenoid at a point 1.50cm from the axis is mathematically represented as
[tex]E_{15} = \frac{ 15*10^{-2 }}{2} * |4.60 |[/tex]
[tex]E_{15} = 0.0345 \ V/m[/tex]
