Question

Number 12 gauge wire, commonly used in household wiring, is 2.053mm in diameter and can safely carry up to 20A. For a wire carrying the maximum current, find the magnetic field strength:

Answer 1

Answer:

The magnetic field strength is 3.9 x 10⁻³ T

Explanation:

The equation for the magnetic field strength produced by a long straight current-carrying wire is written as;

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

Where;

μ₀ is constant = 4π x 10⁻⁷ N/A

I is the maximum current = 20 A

r is the radius of the wire = 2.053/2 = 1.0265 mm = 1.0265 x 10⁻³ m

Substitute in this values into the equation above;

$B = \frac{4 \pi X10^{-7} X 20}{2\pi X 1.0265X10^{-3} } \\\\B =3.9 X10^{-3} T$

Therefore, the magnetic field strength is 3.9 x 10⁻³ T

Answer 2

Number 12 gauge wire, commonly used in household wiring, is 2.053mm in diameter and can safely carry up to 20A. For a wire carrying the maximum current, the magnetic field strength is 3.89 × 10⁻⁶ mT

The magnetic field strength is dependent on the current (I) in the wire and radius (R) which is the distance from the wire.

The magnitude of magnetic field strength can be computed by using the formula;

$\mathbf{B = \dfrac{\mu_oI}{2 \pi R}}$

where;

• magnetic permeability  $\mathbf{\mu_o=4 \pi \times 10^{-7}}$  
• current I = 20 A
• radius R = 2.053 mm/2 = 1.0265 mm

$\mathbf{B = \dfrac{4 \pi \times 10^{-7} \times 20 }{2 \pi \times 1.0265 }}$

B = 3.89 × 10⁻⁶ mT

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