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Otrada [13]
3 years ago
9

A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside

the solenoid near its center?
Physics
1 answer:
RUDIKE [14]3 years ago
8 0

Answer:

Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

Explanation:

Given;

number of turns of solenoid, N = 269 turn

length of the solenoid, L = 102 cm = 1.02 m

radius of the solenoid, r = 2.3 cm = 0.023 m

current in the solenoid, I = 3.9 A

Magnitude of the magnetic field inside the solenoid near its centre is calculated as;

B = \frac{\mu_o NI}{l} \\\\

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T

Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

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