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a solenoid 10.0 cm in diameter and 75.0 cm long is made from copper wire of diameter 0.1 cm,with very thin insulation. the wire is wound onto a cardboard tube in a single layer, with adjacentturns touching each other. what power must be delivered to the solenoid if it is to produce a field of8.00 mt at its center?

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The power required to produce a magnetic field of 8.00 mT at the center of the solenoid is approximately 1.19 × 10¹⁹ W.

To calculate the power required to produce a magnetic field of 8.00 mT at the center of the solenoid, we can use the following formula:

P = I² × R

where:

P is the power in watts (W)

I is the current in amperes (A)

R is the resistance in ohms (Ω)

The current required to produce the desired magnetic field can be calculated using Ampère's law:

B = μ₀ × n × I

where:

B is the magnetic field in teslas (T)

μ₀ is the permeability of free space (4π × 10⁻⁷ T⋅m/A)

n is the number of turns per meter (turns/m)

I is the current in amperes (A)

The resistance of the solenoid can be calculated using the following formula:

R = ρ × L / A

where:

R is the resistance in ohms (Ω)

ρ is the resistivity of copper (1.68 × 10⁻⁸ Ω·m)

L is the length of the solenoid in meters (m)

A is the cross-sectional area of the wire in square meters (m²)

Plugging in the values, we get:

P = (8.00 × 10⁻³ T)² × (4π × 10⁻⁷ T·m/A)² × (750 turns / 0.75 m) / (1.68 × 10⁻⁸ Ω·m) × (π × (0.05 m)²)) ≈ 1.19 × 10¹⁹ W

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