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a long solenoid of 350 turns and length 28.0 cm located in vacuum carries a current of 22.0 ma. the radius of the solenoid is 7.50 cm. a single conducting loop of radius 10.0 cm is placed concentric with the solenoid at its center. determine the magnitude and direction of the current that makes the magnetic field at the center of the solenoid zero

User APerson
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The magnitude of the current in the loop is 40.8 A and the direction of the current in the loop is such that it produces a magnetic field that is in the opposite direction of the magnetic field produced by the solenoid.

To determine the magnitude and direction of the current that makes the magnetic field at the center of the solenoid zero, we can use the principle of superposition. The magnetic field at the center of the solenoid is the sum of the magnetic fields produced by each of the turns of the solenoid. The magnetic field produced by a single turn is given by Ampère's law:

B = μ₀I/2πa

where:

B is the magnetic field

μ₀ is the permeability of free space

I is the current in the turn

a is the distance from the center of the turn to the point where the magnetic field is being measured

In this case, I = 22.0 mA, a = 7.50 cm, and μ₀ = 4π × 10⁻⁷ T⋅m/A. The magnetic field produced by a single turn is therefore:

B = (4π × 10⁻⁷ T⋅m/A) × (22.0 × 10⁻³ A) / (2π × 0.0750 m) ≈ 1.47 × 10⁻⁴ T

The magnetic field produced by each turn is in the same direction, so the magnetic fields produced by all of the turns add up. The total magnetic field at the center of the solenoid is therefore:

B_total = N × B = 350 × 1.47 × 10⁻⁴ T ≈ 0.515 T

We want the magnetic field at the center of the solenoid to be zero, so we need to add a second magnetic field that cancels out the first one. The magnetic field produced by a single loop is given by:

B = μ₀NI/2πr

where:

B is the magnetic field

μ₀ is the permeability of free space

N is the number of turns in the loop

I is the current in the loop

r is the distance from the center of the loop to the point where the magnetic field is being measured

In this case, N = 1, r = 10.0 cm, and μ₀ = 4π × 10⁻⁷ T⋅m/A. The magnetic field produced by the loop is therefore:

B = (4π × 10⁻⁷ T⋅m/A) × 1 × I / (2π × 0.100 m) ≈ 2.00 × 10⁻⁵ T

The magnetic field produced by the loop is in the opposite direction of the magnetic field produced by the solenoid, so we can add the two magnetic fields to get the total magnetic field at the center of the solenoid. The total magnetic field is therefore:

B_total = B_solenoid + B_loop = 0.515 T - 2.00 × 10⁻⁵ T ≈ 0.513 T

The magnitude of the current in the loop is therefore:

I = (2πr/μ₀) × B × N ≈ (2π × 0.100 m / (4π × 10⁻⁷ T⋅m/A)) × 0.513 T × 1 ≈ 40.8 A

User Swashata Ghosh
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