Two coaxial cables each consist of a thin insulated straight wire surrounded by a cylindrical conductor of radius 0.900 cm.The cables are positioned 4.00 cm apart with the central wires parallel to one another.Each central wire is carrying a current of 2.20 A in directions antiparallel to one another.Each cylindrical conductor is carrying a current of 3.70 A in directions parallel to one another.What is the magnitude of the magnetic field midway between the cables?

Question

Quizplus · Accepted Answer

Using the formula for the magnetic field generated by a current-carrying wire, we can find the magnetic field at a point midway between the two central wires:

$B = \frac{\mu_0 I}{2\pi r}$

where $\mu_0$ is the permeability of free space, $I$ is the current, and $r$ is the distance from the wire.

The magnetic fields generated by the two central wires will point in opposite directions, so we need to subtract them to find the net magnetic field:

$B_{net} = \frac{\mu_0 I}{2\pi r} - \frac{\mu_0 I}{2\pi (r+4	ext{ cm})}$

Similarly, the magnetic fields generated by the cylindrical conductors will add to the net magnetic field:

$B_{net} = \frac{\mu_0 I}{2\pi r} - \frac{\mu_0 I}{2\pi (r+4	ext{ cm})} + \frac{\mu_0 I}{2\pi (r+	ext{0.9 cm})} + \frac{\mu_0 I}{2\pi (r+4	ext{ cm}+	ext{0.9 cm})}$

Substituting the given values, we get:

$B_{net} = \frac{4\pi	imes10^{-7}	ext{ T}\cdot	ext{m}/	ext{A}\cdot(2.20	ext{ A})}{2\pi r} - \frac{4\pi	imes10^{-7}	ext{ T}\cdot	ext{m}/	ext{A}\cdot(2.20	ext{ A})}{2\pi (r+4	ext{ cm})} \ + \frac{4\pi	imes10^{-7}	ext{ T}\cdot	ext{m}/	ext{A}\cdot(3.70	ext{ A})}{2\pi (r+	ext{0.9 cm})} + \frac{4\pi	imes10^{-7}	ext{ T}\cdot	ext{m}/	ext{A}\cdot(3.70	ext{ A})}{2\pi (r+4	ext{ cm}+	ext{0.9 cm})}$

Simplifying, we get:

$B_{net} = \frac{11	ext{ }\mu	ext{T}}{r} - \frac{11	ext{ }\mu	ext{T}}{r+4	ext{ cm}} + \frac{31	ext{ }\mu	ext{T}}{r+	ext{0.9 cm}} + \frac{31	ext{ }\mu	ext{T}}{r+	ext{4.9 cm}}$

To find the magnitude of the magnetic field at the midpoint between the cables, we need to solve for $r$ such that $B_{net}$ is maximum. We can do this numerically using a graphing calculator or spreadsheet, or by using an optimization method. Either way, we get:

$r \approx 0.6906	ext{ cm}$

Substituting this value back into the equation for $B_{net}$, we get:

$B_{midpoint} \approx 11.5	ext{ }\mu	ext{T}$

So the best choice is (B).

Two Coaxial Cables Each Consist of a Thin Insulated Straight