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The magnetic field 36.0 cm away from a long, straight wire carrying current 5.00 A is 2780nT.

(a) At what distance is it 278nT ?

User Jorgeca
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1 Answer

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Final answer:

The distance at which the magnetic field is 278nT is 3.61 cm. The magnetic field 36.0cm away from the middle of the straight cord, in the plane of the two wires, is 33.3μT. The distance at which the magnetic field is one-tenth as large is 3.60cm. The magnetic field outside the coaxial cable is zero because the net current is zero.

Step-by-step explanation:

(a) To find the distance at which the magnetic field is 278nT, we can set up a proportion:

2780nT / 36.0cm = 278nT / x

Cross multiplying, we get:

2780nT * x = 278nT * 36.0cm

Simplifying:

x = (278nT * 36.0cm) / 2780nT

Calculating this gives us x = 3.61cm.

(b) To find the magnetic field 36.0cm away from the middle of the straight cord, in the plane of the two wires, we can use the formula for the magnetic field produced by a long, straight wire:

B = (μ₀ * I) / (2π * r)

Plugging in the values, we have:

B = (4π × 10^-7 Tm/A * 5.00A) / (2π * 0.03m)

Simplifying this gives us B = 33.3μT.

(c) To find the distance at which the magnetic field is one-tenth as large, we can set up a proportion:

2780nT / 36.0cm = (2780nT / 10) / x

Cross multiplying, we get:

2780nT * x = (2780nT / 10) * 36.0cm

Simplifying:

x = (2780nT / 10) * 36.0cm / 2780nT

Calculating this gives us x = 3.60cm.

(d) The magnetic field outside the coaxial cable is given by:

B = (μ₀/2π) * (I / r)

Since the currents in the center wire and the sheath are equal and opposite, the net current is zero and the magnetic field outside the cable is zero.

User Ahmed Eid
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