Question

20. The magnetic field inside a 5.0-cm-diameter solenoid is$2.0\text{T}$ and decreasing at $4.0\text{T}/\text{s}$. What is the electric field strength inside the solenoid at a point (a) on the axis and (b) $2.0\text{cm}$from the axis?

Short answer

Part (a) Electric field strength inside the solenoid at a point on axis is$0\text{V}/\text{m}$

Part (b) Electric field strength inside the solenoid at a point$2\text{cm}$ away from the axis is$0.04\text{V}/\text{m}$

Step-by-step solution

Step 1. Given information

Diameter of solenoid$=5.0\text{cm}$

Radius of solenoid$=2.5\text{cm}$

Rate of changing magnetic field$=4\text{T}/\text{s}$

Conversion of units

$\frac{5.0\text{cm}}{1}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}=5.0\times {10}^{-2}\text{m}$

Part (a)

The induced electric field in a region where the radius of circle is less than radius of circular region the charge is confined is given by $E=\frac{r}{2}\left|\frac{dB}{dt}\right|$

Here, $\frac{dB}{dt}$is the rate of changing magnetic field, $r$is the radius of the solenoid.

When the point is on the axis, the radius (r) is zero, therefore,$r=0$

Substituting the value of changing magnetic field and radius in above equation, the induced magnetic field strength is given by

$E=\frac{0}{2}|4\text{T}/\text{s}|E=0\text{V}/\text{m}$

Electric field strength inside the solenoid at a point on axis is $0\text{V}/\text{m}$

Part (b)

When the point is away from the axis, therefore, $r=2.0\text{cm}$

The induced electric field is given by

$E=\frac{r}{2}\left|\frac{dB}{dt}\right|$

Substituting the value of changing magnetic field and radius in above equation, the induced magnetic field strength is given by

$\begin{array}{l}E=\frac{0.02}{2}(4\text{T}/\text{s})\\ E=0.04\text{V}/\text{m}\end{array}$

Electric field strength inside the solenoid at a point$2\text{cm}$ away from the axis is $0.04\text{V}/\text{m}$