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Answer:
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