A 28-turn solenoid of length 6.4 cm produces a magnetic field of magnitude 2 mT at its center. Find the current in the solenoid.
Question
A 28-turn solenoid of length 6.4 cm produces a magnetic field of magnitude 2 mT at its center. Find the current in the solenoid.
Solution
To find the current in the solenoid, we can use Ampere's law which states that 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 (4π x 10^-7 T m/A)
- n is the number of turns per unit length (turns/meter)
- I is the current
We are given:
- B = 2 mT = 2 x 10^-3 T (converting milliTesla to Tesla)
- The number of turns = 28
- The length of the solenoid = 6.4 cm = 6.4 x 10^-2 m (converting cm to m)
First, we need to find n, the number of turns per unit length:
n = Number of turns / Length n = 28 / (6.4 x 10^-2) n = 437.5 turns/meter
Now we can substitute B, μ₀, and n into the formula and solve for I:
2 x 10^-3 = 4π x 10^-7 * 437.5 * I I = 2 x 10^-3 / (4π x 10^-7 * 437.5) I = 1.15 A
So, the current in the solenoid is approximately 1.15 Amperes.
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