A long solenoid has a radius of 2.0 cm and has 700 turns/m. If the current in the solenoid is decreasing at the rate of 8.0 A/s, what is the magnitude of the induced electric field at a point 2.5 cm from the axis of the solenoid?

Question

Quizplus · Accepted Answer

The magnitude of the induced electric field at a point 2.5 cm from the axis of the solenoid is given by:$$E = \frac{\mu_0 N}{2\pi r}\frac{dI}{dt}$$where:* $\mu_0$ is the permeability of free space ($4\pi 	imes 10^{-7} 	ext{ T}\cdot	ext{m/A}$)* $N$ is the number of turns per unit length (700 turns/m)* $r$ is the distance from the axis of the solenoid (2.5 cm = 0.025 m)* $dI/dt$ is the rate of change of current (-8.0 A/s)Plugging in the values, we get:$$E = \frac{(4\pi 	imes 10^{-7} 	ext{ T}\cdot	ext{m/A})(700 	ext{ turns/m})}{2\pi (0.025 	ext{ m})}(-8.0 	ext{ A/s}) = 56 \mu	ext{V/m}$$Therefore, the magnitude of the induced electric field is 56 μV/m.

A Long Solenoid Has a Radius of 2