A long solenoid has a radius of 4.0 cm and has 800 turns/m.If the current in the solenoid is increasing at the rate of 3.0 A/s,what is the magnitude of the induced electric field at a point 2.2 cm from the axis of the solenoid? A)3.3 *10 ^- ⁵ V/m B)3.6 * 10 ^- ⁵ V/m C)3.9 *10 ^- ⁵ V/m D)4.2 *10 ^- ⁵ V/m E)6.0 *10 ^- ⁵ V/m

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Quizplus · Accepted Answer

The magnitude of the induced electric field at a point 2.2 cm from the axis of the solenoid is given by:$$E = \frac{d\Phi_B}{dt} = \frac{d}{dt}(\mu_0 n I A)$$where:* $d\Phi_B/dt$ is the rate of change of magnetic flux* $\mu_0$ is the permeability of free space* $n$ is the number of turns per unit length* $I$ is the current* $A$ is the area of the solenoidSubstituting the given values into the equation, we get:$$E = \frac{d}{dt}(\mu_0 n I A) = \mu_0 n \frac{dI}{dt} A = (4\pi 	imes 10^{-7} 	ext{ T$\cdot$m/A})(800 	ext{ turns/m})(3.0 	ext{ A/s})(\pi (0.022 	ext{ m})^2) = 3.3 	imes 10^{-5} 	ext{ V/m}$$Therefore, the magnitude of the induced electric field at a point 2.2 cm from the axis of the solenoid is 3.3 x 10^-5 V/m.

A Long Solenoid Has a Radius of 4