Question

A long solenoid carries 1( current. The diameter of wire, which used for its construction is 2 mm, then magnetic field at the centre of solenoid :​

Answer 1

Explanation:

The turns of wire on the solenoid are shown in figure.

Number of turns in 1mm =1

n=1000turns/m

∴B=μ  

0

​

ni

=4π×10  

−7

×10  

3

×5

=2π×10  

−3

T.

Answer 2

Answer: The magnetic field at the center of the solenoid is approximately 0.9 mT.

The magnetic field inside a long solenoid can be approximated by a uniform field, given by the equation:

                               B = μ₀ * n * I

where B is the magnetic field, μ₀ is the permeability of free space (4π x 10^-7 T·m/A), n is the number of turns per unit length, and I is the current.

In this case, the solenoid carries a current of 1 A, and the diameter of the wire used for its construction is 2 mm. Assuming the wire is wound tightly and uniformly, we can calculate the number of turns per unit length (n) as follows:

                             n = N / L

where N is the total number of turns and L is the length of the solenoid. Since the solenoid is long, we can assume that the length is much greater than the diameter, so we can neglect the end effects and use the equation:

                             L = π * d * N

where d is the diameter of the solenoid (not the wire). Since the diameter of the wire is given as 2 mm, we can assume that the diameter of the solenoid is slightly larger, say 2.2 mm, to account for the thickness of the insulation.

Therefore, we have:

                            d = 2.2 mm = 0.0022 m

                            N = ?

                           L = π * 0.0022 * N

To find the total number of turns (N), we need to know the length of the solenoid. Let's assume that the solenoid is 1 meter long (which is much greater than the diameter of the solenoid).

Then we have:

                           L = π * 0.0022 * N = 1 m

Solving for N, we get:

                          N = 1 / (π * 0.0022) = 721.18

Now we can calculate the magnetic field at the center of the solenoid using the equation:

                                 B = μ₀ * n * I

where μ₀ is the permeability of free space (4π x 10^-7 T·m/A), n is the number of turns per unit length (N/L), and I is the current (1 A).

Substituting the values, we get:

                    B = 4π x 10^-7 * 721.18 / 1 * 1 = 9.0 x 10^-4 T

Therefore, the magnetic field at the center of the solenoid is approximately 0.9 mT.

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