Question

A long solenoid (radius = 3.0 cm, 2500 turns per meter) carries a current given by I = 0.30 sin(200 t) A, where t is measured in s. When t = 2.5 ms, what is the magnitude of the induced electric field at a point which is 4.0 cm from the axis of the solenoid? The answer is 1.9*10^-3 V/m. please explain in detail

Answer 1

Answer:

1.86 × 10 ∧-3 V/m

≅ 1.9 × 10 ∧-3 V/m

Explanation:

Radius of long solenoid (R)= 3 cm

Turns per meter (n) = 2500

current flowing in coil (I) = 0.30 sin (200 t)

Time (t) = 2.5 ms

Distance at which electric field calculated (r) = 4 cm

The main solution is shown in the picture attached below.

Thank you and i hope it helps.

Answer 2

The magnitude of the induced electric field at a point, 4.0 cm from the axis of the solenoid is $\bold {\vec E = 1.9x10^{-3}\ V/m}$.

Given Here,

Raadius of solenoid = R

Turns per meter = 2500

Current, I =  0.30 sin(200 t) A,

Time,  t = 2.5 ms,

Distance at which electric field = 4 cm

From Ampere law,

$\int\limits { \vec {E}} \, d\vec s = \dfrac {d }{dt} (M_0 AI)$

Put the values in the formula,

$\bold {\dfrac {dI}{dt} = \dfrac {d (0.3 sin (200t))}{dt}}$

Solving it, we get,

$\bold {\dfrac {dI}{dt} = 52.65}$

Put the value in equation 1 and solve for E, we get,

$\bold {\vec E = 1.9x10^{-3}\ V/m}$

Therefore, the magnitude of the induced electric field at a point, 4.0 cm from the axis of the solenoid is $\bold {\vec E = 1.9x10^{-3}\ V/m}$.

To know more about solenoid,

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