Nonnuclear submarines use batteries for power when submerged.
Find the magnetic field 55 cm from a straight wire carrying 1150 A from the batteries to the drive mechanism of a submarine.
The Correct Answer and Explanation is:
The magnetic field a distance rr from a long, straight current‑carrying conductor is given by Ampère’s law (or the Biot–Savart law in its simplified form for straight wires): B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}
where
- BB is the magnetic field (in teslas, T),
- μ0\mu_0 is the permeability of free space (4π×10−7 T⋅m/A)\bigl(4\pi\times10^{-7}\,\mathrm{T\cdot m/A}\bigr),
- II is the current through the wire (in amperes, A),
- rr is the perpendicular distance from the wire (in meters, m).
Calculation
For I=1150 AI = 1150\ \mathrm{A} and r=0.55 mr = 0.55\ \mathrm{m}: B=(4π×10−7 T⋅m/A)×1150 A2π×0.55 m=4π×10−7×11502π×0.55 T=2×1150×10−70.55 T=2.30×10−40.55 T≈4.18×10−4 T.\begin{aligned} B &= \frac{(4\pi\times10^{-7}\,\mathrm{T\cdot m/A}) \times 1150\ \mathrm{A}}{2\pi \times 0.55\ \mathrm{m}} \\[6pt] &= \frac{4\pi\times10^{-7} \times 1150}{2\pi \times 0.55}\ \mathrm{T} \\[6pt] &= \frac{2 \times 1150 \times 10^{-7}}{0.55}\ \mathrm{T} \\[6pt] &= \frac{2.30 \times 10^{-4}}{0.55}\ \mathrm{T} \\[6pt] &\approx 4.18\times10^{-4}\ \mathrm{T}. \end{aligned}
Thus B≈4.18×10−4 tesla=418 μT.\boxed{B \approx 4.18\times10^{-4}\ \text{tesla} = 418\ \mu\text{T}.}
Explanation and Context
- Origin via Ampère’s Law
Ampère’s law in integral form states ∮B⋅dℓ=μ0Ienc\displaystyle \oint \mathbf{B}\cdot d\boldsymbol\ell = \mu_0 I_{\rm enc}. For a circular path of radius rr concentric with a straight wire, B\mathbf{B} is tangent and constant in magnitude, so B(2πr)=μ0IB (2\pi r)=\mu_0 I, yielding the formula above. - Role in Submarine Systems
In non‑nuclear submarines, high‑capacity batteries supply large currents (on the order of 10³ A) to electric motors that drive the propellers when submerged. The returning current through the hull or dedicated return conductor thus generates magnetic fields in the vicinity of these power cables. - Relative Strength
A field of 4.18×10−4 T4.18\times10^{-4}\,\mathrm{T} is about 7–8 times stronger than Earth’s magnetic field (~50 μT). In submarine stealth considerations, such local magnetic anomalies can be detected by magnetic anomaly detectors (MAD) in anti‑submarine aircraft, so power‑cable routing and magnetic signature reduction techniques (e.g., degaussing coils) are critical. - Direction and Right‑Hand Rule
The magnetic field lines circle the wire according to the right‑hand rule: if the thumb points in the direction of current flow, the curled fingers show the sense of B\mathbf{B}. At a point fixed relative to the wire, reversing the current reverses the field direction. - Distance Dependence
Note the 1/r1/r dependence: doubling the distance to 1.10 m1.10\ \mathrm{m} would halve the field to ∼2.09×10−4 T\sim2.09\times10^{-4}\,\mathrm{T}. This rapid spatial decay underscores why shielding and physical separation can mitigate stray magnetic fields in sensitive onboard equipment.
Answer: B≈4.18×10−4 TB \approx 4.18\times10^{-4}\ \mathrm{T}