To find the strength of the magnetic field created by a current carrying wire, you will need to know the current and how far away from the wire the point in question is. You will also use the right-hand rule for current-carrying wire to determine the direction of the magnetic field.
Strength of a Magnetic Field Around a Current-Carrying Wire
\(\mathsf{ B = \frac{\mu_0 I}{2 \pi r} }\)
...where B is the magnetic field strength, μ0 is 4π x 10-7 Tm/A, I is the current in the wire, and r is the distance the reference point is away from the wire.
Use the equation above and the right-hand rule to answer the following questions. Work through them on your own and then check your answers against the answers at the end.
A current-carrying wire carries a current of 0.545 A. What is the magnitude and direction of the magnetic field at point P, 0.034 meters away from the wire?
A current-carrying wire carries a current of 2.10 A. What is the magnitude and direction of the magnetic field at point P, 0.00520 meters away from the wire?
A current-carrying wire carries a current of 1.85 A. How far away from the wire is point P if the magnetic field at that location is 7.40 x 10-6 T?
| Your Responses | Sample Answers |
|---|---|
| \(\mathsf{ B = \frac{4 \pi \times 10^{-7} \text{ Tm/A} \cdot 0.545 \text{ A}}{2 \pi \cdot 0.034 \text{ m}} }\) \(\mathsf{ B = 3.21 \times 10^{-6} \text{ T} }\) At point P, the field is coming out of the page. |
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| \(\mathsf{ B = \frac{4 \pi \times 10^{-7} \text{ Tm/A} \cdot 2.10 \text{ A}}{2 \pi \cdot 0.00520 \text{ m}} }\) \(\mathsf{ B = 8.08 \times 10^{-5} \text{ T} }\) At point P, the field is coming out of the page. |
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| \(\mathsf{ 7.40 \times 10^{-6} \text{ T} = \frac{4 \pi \times 10^{-7} \text{ Tm/A} \cdot 1.85 \text{ A}}{2 \pi r} }\) \(\mathsf{ r = \frac{4 \pi \times 10^{-7} \text{ Tm/A} \cdot 1.85 \text{ A}}{2 \pi \cdot 7.40 \times 10^{-6} \text{ T}} }\) \(\mathsf{ r = 0.05 \text{ m} }\) |
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