A 2.0-m lamp cord leads from the 110-V outlet to a lamp having a 75-W lightbulb. The cord consists of two insulated parallel wires 8.0 mm apart and held together by the insulation. One wire carries the current into the bulb, and the other carries it out.

1. What is the magnitude of the magnetic field the cord produces midway between the two wires?

2. What is the magnitude of the magnetic field the cord produces 2.0 mm from one of the wires in the same plane in which the two wires lie?

3. Compare the field between the two wires with Earth’s magnetic field (0.5×10⁻⁴ T). Give your answer as B/BEarth.

a) Magnitude of the magnetic field midway between the two wires = 0.0000682 T = (6.82 × 10⁻⁵) T

b) Magnitude of the magnetic field the cord produces 2.0 mm from one of the wires in the same plane in which the two wires lie

= 0.00009093 T = (9.093 × 10⁻⁵) T

c) B(earth) = (0.5×10⁻⁴) T

B(a) = Magnetic field midway between the wires = (6.82 × 10⁻⁵) T

B(b) = Magnetic field at a point 2 mm from one of the wires = (9.093 × 10⁻⁵) T

Both magnetic fields exceed the earth's magnetic field.

(Ba/Bearth) = 1.364

(Bb/Bearth) = 1.8186

Explanation:

First of we obtain the current in the wires.

P = IV

I = (P/V) = (75/110) = 0.682 A

Magnetic field at a distance from a current carrying wire is given as

B = μ₀I/2πr

if two wires carry current in opposite direction, the magnetic field produced between them is the sum of the individual magnetic field due to each wire.

B = [μ₀I₁/2πr] + [μ₀I₂/2πr]

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

I₁ = current in wire 1 = 0.682 A

I₂ = current in wire 2 = 0.682 A

r = distance at which we want to evaluate the magnetic field due to the current in the wires = (8mm/2) = 4 mm = 0.004 m

I₁ = I₂ = 0.682 A

B = 2[μ₀I/2πr] = [μ₀I₁/πr]

B = (4π × 10⁻⁷× 0.682)/(π×0.004)

B = 0.0000682 T = (6.82 × 10⁻⁵) T

b) magnitude of the magnetic field the cord produces 2.0 mm from one of the wires in the same plane in which the two wires lie

B = [μ₀I₁/2πr₁] + [μ₀I₂/2πr₂]

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

I₁ = current in wire 1 = 0.682 A

I₂ = current in wire 2 = 0.682 A

r₁ = distance at which we want to evaluate the magnetic field due to current in wire 1 = 2mm = 0.002 m

r₂ = distance at which we want to evaluate the magnetic field due to the current in the wire 2 = 8 mm - 2 mm = 6 mm = 0.006 m

I₁ = I₂ = 0.682 A

B = [μ₀I/2π] [(1/r₁) + (1/r₂)]

B = [4π × 10⁻⁷ × 0.682/2π] [(1/0.002) + (1/0.006)]

B = (1.364 × 10⁻⁷) [666.6667] = 0.00009093 T = (9.093 × 10⁻⁵) T

c) Compare the field between the two wires with Earth’s magnetic field (0.5×10⁻⁴ T). Give your answer as B/BEarth.

The two fields obtained in (a) and (b) above, are both more than the earth's magnetic field.

B(earth) = (0.5×10⁻⁴) T

B(a) = Magnetic field midway between the wires = (6.82 × 10⁻⁵) T

B(b) = Magnetic field at a point 2 mm from one of the wires = (9.093 × 10⁻⁵) T

(Ba/Bearth) = (6.82 × 10⁻⁵)/(0.5×10⁻⁴) = 1.364

(Bb/Bearth) = (9.093 × 10⁻⁵)/(0.5×10⁻⁴) = 1.8186

Hope this Helps!!!

electrical power

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.