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A solenoid 25.0 cm long and with a cross-sectional area of 0.500

cm2 contains 400 turns of wire and carries a current of 80.0 A.
Calculate: (a) the magnetic field in the solenoid

User Joran
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1 Answer

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Final answer:

The magnetic field inside a solenoid is calculated using the formula B = μ_0 * (n * I), where μ_0 is the permeability of free space, n is the number of turns per unit length, and I is the current. The magnetic field in a 25.0 cm long solenoid with 400 turns and 80.0 A current is 1.60 x 10^-2 T.

Step-by-step explanation:

To calculate the magnetic field inside a solenoid, we use Ampère's Law which simplifies to the formula:

B = μ_0 * (n * I)

Where B is the magnetic field, μ_0 is the permeability of free space (μ_0 = 4π x 10^-7 T*m/A), n is the number of turns per unit length of the solenoid, and I is the current passing through the coils.

The number of turns per unit length n is calculated as the total number of turns divided by the length of the solenoid:

n = Total turns / Length

For a solenoid that is 25.0 cm long with 400 turns and current of 80.0 A, the magnetic field can be calculated as follows:

n = 400 turns / 0.25 m = 1600 turns/m

Then, plug into the formula to find B:

B = (4π x 10^-7 T*m/A) * (1600 turns/m * 80.0 A)

B = 1.60 x 10^-2 T

Therefore, the magnetic field in the solenoid is 1.60 x 10^-2 T (Tesla).

User Ziya
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