Question

A time dependent magnetic field inside a long solenoid of radius 0.05m is given by \(\overrightarrow𝐵\)(𝑡) = 𝐵₀ sin 𝜔𝑡 \(\^z\). If 𝜔=100rad/s and 𝐵₀= 0.98Weber/m² , then the amplitude of the induced electric field at a distance of 0.07m from the axis of the solenoid is ______V/m. (Rounded off to two decimal places)

Answer 1

Correct Answer: 1.71

Induced Electric Field in a Solenoid

The induced electric field \( E(r) \) at a distance \( r \) from the axis of a solenoid is obtained from Faraday’s law of electromagnetic induction:

\[ \oint \mathbf{E} \cdot d\mathbf{l} = -\frac{d\Phi_B}{dt} \]

where \( \Phi_B \) is the magnetic flux through the solenoid.

1. Case 1: \( r < R \)

The magnetic flux through a circle of radius \( r \) inside the solenoid is:

\[ \Phi_B = B(t) \cdot \pi r^2 \]

2. Case 2: \( r > R \)

The magnetic flux through the solenoid remains constant and is determined by its radius \( R \):

\[ \Phi_B = B(t) \cdot \pi R^2 \]

Step 1: Expression for the Induced Electric Field

For \( r = 0.07m > R =0.05m \), the induced electric field \( E(r) \) is given by:

\[ E(r) \cdot 2\pi r = -\frac{d}{dt} \left( B(t) \cdot \pi R^2 \right) \]

Step 2: Substituting \( B(t) \)

Substituting \( B(t) = B_0 \sin(\omega t) \), the time derivative is:

\[ \frac{dB}{dt} = B_0 \omega \cos(\omega t) \]

Thus, the induced electric field is:

\[ E(r) = \frac{R^2}{2r} \cdot B_0 \omega \]

Step 3: Substituting Given Values

• \( R = 0.05m \)
• \( r = 0.07m \)
• \( B_0 = 0.98 \) Weber/m²
• \( \omega = 100 \) rad/s

\[ E(r) = \frac{(0.05)^2}{2 \times 0.07} \times 0.98 \times 100 \]

Step 4: Simplifying the Expression

\[ E(r) = \frac{0.0025}{0.14} \times 98 \]

\[ E(r) = 1.71 \text{ V/m} \]

Final Answer

The amplitude of the induced electric field is 1.71 V/m.