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VashaNatasha
2 months ago
10

A cylindrical insulated wire of diameter 5.0 mm is tightly wound 200 times around a cylindrical core to form a solenoid with adj

acent coils touching each other. When a 0.10 A current is sent through the wire, what is the magnitude of the magnetic field on the axis of the solenoid near its center
Physics
2 answers:
Yuliya22 [3.3K]2 months ago
7 0

Answer:

B = 0.0016 T

Explanation:

Provided information:

- Wire diameter, dw = 5.0 mm

- Coil turns, N = 200

- Current in the wire, I = 0.10 A

- Magnetic permeability of free space, μ₀ = 1.257 x 10^-6 H/m

Objective:

Determine the magnitude of the magnetic field along the axis of the solenoid towards its center.

Solution:

- The magnetic field (B) at the center of the coil, proportional to the number of turns (N) and the current (I) flowing through the wire, can be established through the Biot-Savart Law:

B = μ₀ * (N/L) * I

- Where, L: The length of the coil

- The circumference of a cylindrical core corresponds to the length of coil for one complete turn:

L = π * dw

- Substituting into the Biot-Savart expression:

B = μ₀ * N * I / (π * dw)

B = (4π^-7) * (0.10) * (200) / (π * 0.005)

B = 0.0016 T

kicyunya [3.2K]2 months ago
3 0

Answer:

0.0016 T

Explanation:

Given parameters:

Wire diameter = 5 mm = 0.005 m

Radius of the wire, R = 0.0025 m

Turns of wire, N = 200

Current passing through the wire, I = 0.10A

The expression for the magnitude of the magnetic field is:

B = (μ₀NI) / (2πR)

Where μ₀ is the magnetic permeability in a vacuum.

B = (1.257 * 10⁻⁶ * 200 * 0.1) / (2 * π * 0.0025)

B = 0.0016 T

The magnetic field strength is 0.0016 T.

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