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

The magnitude of the induced electric field at a point 2.5 cm from the axis of the solenoid is 8.8 x 10⁻⁵ V/m

Explanation:

given information:

radius, r = 2.0 cm

N = 700 turns/m

decreasing rate, dI/dt = 9.0 A/s

the magnitude of the induced electric field at a point 2.5 cm (r = 2.5 cm = 0.025 m) from the axis of the solenoid?

the magnetic field at the center of solenoid

B = μ₀nI

where

B = magnetic field (T)

μ₀ = permeability (1.26× 10⁻⁶ T.m/A)

n = the number turn per unit length (turn/m)

I = current (A)

dB/dt = μ₀n dI/dt                                           (1)

now we calculate the induced electric field by using

E =  

= 2E/r                                                     (2)

where

E = the induced electric field (V/m)

we substitute the firs and second equation, thus

dB/dt = μ₀n dI/dt  

2E/r = μ₀n dI/dt  

E = (1/2) r μ₀n dI/dt

  = (1/2) (0.025) (1.26× 10⁻⁶) (700) (8)

  = 8.8 x 10⁻⁵ V/m

it has to be b, about 97 percent.

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