Respuesta :
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.
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
