Question

85. III A 2.0-cm-diameter solenoid is wrapped with 1000 turns per CALC meter. 0.50 cm from the axis, the strength of an induced electric field is$5.0\times {10}^{-4}\text{V}/\text{m}$ . What is the rate $dI/dt$with which the current through the solenoid is changing?

Short answer

The change in current over the change in time is$160\text{A}/\text{s}$

Step-by-step solution

Step 1. given information

Diameter of the solenoid,$D=2\text{cm}=0.02\text{m}$ , number of turns per meter is$n=1000$ , distance $d=0.5\text{cm}=0.005\text{m}$from the axis, the strength of induced electric field is$E=5\times {10}^{-4}\text{V}/\text{m}.$

Step 2. Explanation

The equation for change in current over the change in time,$\frac{dI}{dt}$ .

$\frac{dI}{dt}=\frac{E2\mathcal{l}}{r{\mu }_{0}N}$

Convert from to meters.

$r=(0.50\text{cm})\left(\frac{1\text{m}}{100\text{cm}}\right)=0.0050\text{m}$

Calculate the rate of change of change of current with respect to change in time.

$\frac{dI}{dt}=\frac{E2\mathcal{l}}{r{\mu }_{0}N}$

Substitute$5.0\times {10}^{-4}\text{V}/\text{m}$ for$E,0.0050\text{m}$ for$r$ , and 1000 for turns per length$\frac{N}{l}$ .

$\begin{array}{l}\frac{dI}{dt}=\frac{2E}{r{\mu }_0\left(\frac{N}{\mathcal{l}}\right)}\\ =\frac{2\left(5.0\times {10}^{-4}\text{V}/\text{m}\right)}{(0.0050\text{m})\left(4\pi \times {10}^{-7}\text{Tm}/\text{A}\right)\left(1000{\text{m}}^{-1}\right)}\\ =160\text{A}/\text{s}\end{array}$

Thus, the change in current over the change in time is $160\text{A}/\text{s}$.