A 0.5 m long solenoid has 500 turns and has a flux density of 2.512 X 10-3 T atits center. Find the current in solenoid. Given μ0 = 4π X 10 -7 H/m ,Π =3.14

The magnetic field B inside a solenoid is given by the formula B = μ0 * n * I, where:

• B is the magnetic field,
• μ0 is the permeability of free space,
• n is the number of turns per unit length,
• I is the current.

We can rearrange this formula to solve for I: I = B / (μ0 * n).

Given in the problem, we have:

• B = 2.512 x 10^-3 T,
• μ0 = 4π x 10^-7 H/m,
• The solenoid is 0.5 m long and has 500 turns, so n = 500 turns / 0.5 m = 1000 turns/m.

Substituting these values into the formula gives:

I = (2.512 x 10^-3 T) / [(4π x 10^-7 H/m) * (1000 turns/m)] = (2.512 x 10^-3 T) / (4 * 3.14 * 10^-7 H/m * 1000 turns/m) = (2.512 x 10^-3 T) / (1.256 x 10^-3 H * turns) = 2 A.

So, the current in the solenoid is 2 A.