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The magnetic field b inside a long solenoid carrying a current of 10 ampere is 3.14 * 10⁻² t find the number of turns per unit length of the solenoid

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Final answer:

The number of turns per unit length of the solenoid is 2.5 * 10^4 turns/m.

Step-by-step explanation:

To find the number of turns per unit length of the solenoid, we can rearrange the formula for the magnetic field inside a solenoid: B = μnI, where B is the magnetic field, μ is the permeability of free space, n is the number of turns per unit length, and I is the current. Rearranging the formula gives us: n = B/(μI). Plugging in the values of B = 3.14 * 10-2 T and I = 10 A, and the permeability of free space μ = 4π * 10-7 T*m/A, we can calculate the number of turns per unit length n.

Using the formula n = B/(μI),

n = (3.14 * 10-2 T) / (4π * 10-7 T*m/A * 10 A)

Calculating this, we get n = 2.5 * 104 turns/m. Therefore, the number of turns per unit length of the solenoid is 2.5 * 104 turns/m.

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