Question

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

Answer 1

Answer:

B = 0.0016 T

Explanation:

Given:-

- The diameter of the wire, dw = 5.0 mm

- The number of turns of coil, N = 200

- The current in the wire, I = 0.10 A

- The permeability of free space, uo = 1.257 x 10^-6 H/m

Find:-

what is the magnitude of the magnetic field on the axis of the solenoid near its center

Solution:-

- The magnetic field (B) is generated at the center of coil and proportional to the number of turns (N) and amount of current that flows through the wire (I) can be determined by Biot-Sovart Law:

                               B = uo * (N/L) * I

- Where,                 L: The length of the coil

- The circumference of a cylindrical core is length of coil for one turn:

                               L = pi*dw

- Substitute into the Biot-Sovart expression:

                              B = uo*N*I / (pi*dw)

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

                              B = 0.0016 T

Answer 2

Answer:

0.0016 T

Explanation:

Parameters given:

Diameter of wire = 5 mm = 0.005 m

Radius of wire, R = 0.0025 m

Number of turns, N = 200

Current through the wire, I = 0.10A

The magnitude of the magnetic field is given as:

B = (u₀NI) / (2πR)

Where u = magnetic permeability of free space.

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

B = 0.0016 T

The magnitude of the Magnetic field is 0.0016 T.