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
Question
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
Solution
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.
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