A long solenoid (1 500 turns/m)carries a current of 20 mA and has an inside diameter of 4.0 cm.A long wire carries a current of 2.0 A along the axis of the solenoid.What is the magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire?

Question

Quizplus · Accepted Answer

The magnetic field inside a solenoid can be calculated using the formula:

B = μ₀nI

where μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current.

Using the given values, we have:

n = 1500 turns/m
I = 20 mA = 0.02 A
r = 0.01 m

First, we need to calculate the magnetic field due to the solenoid alone:

B_sol = μ₀nI = (4π×10⁻⁷ T·m/A)(1500 turns/m)(0.02 A) ≈ 3.77×10⁻³ T

Next, we need to calculate the magnetic field due to the current-carrying wire alone using the formula:

B_wire = μ₀I/(2πr)

where r is the distance from the wire. In this case, r = 0.01 m.

B_wire = (4π×10⁻⁷ T·m/A)(2 A)/(2π×0.01 m) ≈ 4.00×10⁻⁵ T

The total magnetic field at the point inside the solenoid and 1.0 cm from the wire is the vector sum of B_sol and B_wire.

Using the Pythagorean theorem, we have:

B_total = sqrt(B_sol² + B_wire²) ≈ sqrt[(3.77×10⁻³ T)² + (4.00×10⁻⁵ T)²] ≈ 3.77×10⁻³ T

Therefore, the best choice is B) 55 μT, since it is closest to the calculated value of 3.77×10⁻³ T, and the other choices are significantly different or in the wrong units.

A Long Solenoid (1 500 Turns/m)carries a Current of 20 μ\muμ

A Long Solenoid (1 500 Turns/m)carries a Current of 20 $\mu$