A 11.1 cm 11.1 cm long solenoid contains 703 turns 703 turns and carries a current of 4.31 A 4.31 A. What is the strength B B of the magnetic field at the center of this solenoid?

Given Information:

Current = I = 4.31 A

Number of turns = N = 703

Length of solenoid = L = 11.1 cm = 0.11 m

Required Information:

Magnetic field = B = ?

Answer:

Magnetic field = 0.346 T

Step-by-step explanation:

The magnetic field (B) produced in a current (I) carrying long solenoid of length (L) is given by

B = μ₀NI/L

Where μ₀ is the permeability of free space and its value is 4πx10⁻⁷ T. m/A

B = (4πx10⁻⁷*703*4.31) / *0.11

B = 0.0346 T

Therefore, a magnetic field of strength 0.0346 T would be produced at the center of this solenoid.

Given that

Length of solenoid

L=11.1cm = 0.111m

Number of turns

N=703turns

Current

I=4.31A

B=?

Thus, the magnetic field in the core of a solenoid is directly proportional to the product of the current flowing around the solenoid and the number of turns per unit length of the solenoid

B=μo•n•I

Where,

μo is constant and has a value of

μo=4π*10^-7

n is number of turn per unit length

n=N/L

n=703/0.111 = 6333.333 turns/m

I is the current, and it is 4.31A in this case

Then,

B=μo•n•i

B=4π*10^-7*6333.333*4.31

B=0.0343T

B=34.3mT

The magnetic field at the center of the coil is 34.3mT