Respuesta :
Answer:
(A). The electric field strength inside the solenoid at a point on the axis is zero.
(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is [tex]3.75\times10^{-2}\ V/m[/tex].
Explanation:
Given that,
Magnetic field = 2.0 T
Diameter = 5.0 cm
Rate of decreasing in magnetic field = 5.00 T/s
(A). We need to calculate the electric field strength inside the solenoid at a point on the axis
Using formula of electric field inside the solenoid
[tex]E=\dfrac{r}{2}|\dfrac{dB}{dt}|[/tex]
Electric field on the axis of the solenoid
Here, r = 0
[tex]E=\dfrac{0}{2}\times5.00[/tex]
[tex]E = 0[/tex]
The electric field strength inside the solenoid at a point on the axis is zero.
(B). We need to calculate the electric field strength inside the solenoid at a point 1.50 cm from the axis
Using formula of electric field inside the solenoid
[tex]E=\dfrac{r}{2}|\dfrac{dB}{dt}|[/tex]
[tex]E=\dfrac{1.50\times10^{-2}}{2}\times|5.00|[/tex]
[tex]E=0.0375= 3.75\times10^{-2}\ V/m[/tex]
Hence, (A). The electric field strength inside the solenoid at a point on the axis is zero.
(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is [tex]3.75\times10^{-2}\ V/m[/tex].
The electric field strength for the 2 given situations inside the solenoid are;
A) E = 0 V/m
B) E = 0.038 V/m
We are given;
Diameter; D = 5.0 cm
Magnetic field; B = 2.0 T
Rate at which magnetic field is decreasing; dB/dt = 5.00 T/s
A) We want to find the electric field strength inside the solenoid at a point on the axis.
At a point on the axis, the radius is 0.
Now, from faradays law, the formula for electric field strength inside solenoid is given as;
E = [tex]\frac{r}{2} * \frac{dB}{dt}[/tex]
Thus, plugging in relevant values, we have;
E = 0/2 × 5
E = 0 V/m
B) We want to find the electric field strength inside the solenoid at a point 1.5 cm from the axis.
This means that r = 1.5 cm = 0.015 m
Thus;
E = (0.015/2) × 5
E = 0.0375 V/m
Approximating to 3 significant figures gives;
E = 0.038 V/m
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