A closely wound, circular coil with a diameter of 3.00 cm has 690 turns and carries a current of 0.420 A. What is the magnitude of the magnetic field at the center of the coil? What is the magnitude of the magnetic field at a point on the axis of the coil a distance of 7.20 cm from its center?
The correct answer and explanation is:
Let’s break down the problem step by step.
Given:
- Diameter of the coil, d=3.00 cm=0.0300 md = 3.00 \, \text{cm} = 0.0300 \, \text{m}
- Number of turns, N=690N = 690
- Current, I=0.420 AI = 0.420 \, \text{A}
- Distance from coil center along axis, x=7.20 cm=0.0720 mx = 7.20 \, \text{cm} = 0.0720 \, \text{m}
Step 1: Calculate the radius of the coil
r=d2=0.03002=0.0150 mr = \frac{d}{2} = \frac{0.0300}{2} = 0.0150 \, \text{m}
Step 2: Magnetic field at the center of the coil
The formula for the magnetic field at the center of a circular coil with NN turns carrying current II is: Bcenter=μ0NI2rB_{\text{center}} = \frac{\mu_0 N I}{2r}
Where:
- μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A} (permeability of free space)
- N=690N = 690
- I=0.420 AI = 0.420 \, \text{A}
- r=0.0150 mr = 0.0150 \, \text{m}
Plug in the numbers: Bcenter=4π×10−7×690×0.4202×0.0150B_{\text{center}} = \frac{4\pi \times 10^{-7} \times 690 \times 0.420}{2 \times 0.0150}
Calculate numerator: 4π×10−7×690×0.420=4π×10−7×289.8≈(12.566×10−7)×289.8=3.641×10−44\pi \times 10^{-7} \times 690 \times 0.420 = 4\pi \times 10^{-7} \times 289.8 \approx (12.566 \times 10^{-7}) \times 289.8 = 3.641 \times 10^{-4}
Calculate denominator: 2×0.0150=0.03002 \times 0.0150 = 0.0300
Thus, Bcenter=3.641×10−40.0300=0.01214 T=12.14 mTB_{\text{center}} = \frac{3.641 \times 10^{-4}}{0.0300} = 0.01214 \, \text{T} = 12.14 \, \text{mT}
Step 3: Magnetic field at a point on the axis a distance xx from the center
The magnetic field on the axis of a circular coil at distance xx from the center is given by: Baxis=μ0NIr22(r2+x2)3/2B_{\text{axis}} = \frac{\mu_0 N I r^2}{2(r^2 + x^2)^{3/2}}
Substitute values: Baxis=4π×10−7×690×0.420×(0.0150)22×((0.0150)2+(0.0720)2)3/2B_{\text{axis}} = \frac{4\pi \times 10^{-7} \times 690 \times 0.420 \times (0.0150)^2}{2 \times \left((0.0150)^2 + (0.0720)^2\right)^{3/2}}
Calculate r2r^2: r2=(0.0150)2=2.25×10−4r^2 = (0.0150)^2 = 2.25 \times 10^{-4}
Calculate x2x^2: x2=(0.0720)2=5.184×10−3x^2 = (0.0720)^2 = 5.184 \times 10^{-3}
Calculate denominator inside the parentheses: r2+x2=2.25×10−4+5.184×10−3=5.409×10−3r^2 + x^2 = 2.25 \times 10^{-4} + 5.184 \times 10^{-3} = 5.409 \times 10^{-3}
Now calculate (r2+x2)3/2(r^2 + x^2)^{3/2}: (5.409×10−3)3/2=(5.409×10−3)1.5=5.409×10−33(5.409 \times 10^{-3})^{3/2} = (5.409 \times 10^{-3})^{1.5} = \sqrt{5.409 \times 10^{-3}}^3
First, 5.409×10−3=0.07358\sqrt{5.409 \times 10^{-3}} = 0.07358, then cube it: 0.073583=0.0003980.07358^3 = 0.000398
Calculate numerator: 4π×10−7×690×0.420×2.25×10−44\pi \times 10^{-7} \times 690 \times 0.420 \times 2.25 \times 10^{-4}
Calculate stepwise:
- 4π×10−7≈1.2566×10−64\pi \times 10^{-7} \approx 1.2566 \times 10^{-6}
- Multiply by 690: 1.2566×10−6×690=8.671×10−41.2566 \times 10^{-6} \times 690 = 8.671 \times 10^{-4}
- Multiply by 0.420: 8.671×10−4×0.420=3.641×10−48.671 \times 10^{-4} \times 0.420 = 3.641 \times 10^{-4}
- Multiply by 2.25×10−42.25 \times 10^{-4}: 3.641×10−4×2.25×10−4=8.192×10−83.641 \times 10^{-4} \times 2.25 \times 10^{-4} = 8.192 \times 10^{-8}
Denominator is 2×0.000398=0.0007962 \times 0.000398 = 0.000796
Finally, Baxis=8.192×10−80.000796=1.029×10−4 T=0.103 mTB_{\text{axis}} = \frac{8.192 \times 10^{-8}}{0.000796} = 1.029 \times 10^{-4} \, \text{T} = 0.103 \, \text{mT}
Final answers:
- Magnetic field at center: 12.14 mT
- Magnetic field at 7.20 cm on axis: 0.103 mT
Explanation (300 words):
The magnetic field generated by a current-carrying coil depends on the coil’s size, number of turns, and the current passing through it. For a coil, the magnetic field is strongest at its center because all the circular loops contribute constructively there.
At the center, the magnetic field is calculated using the formula B=μ0NI2rB = \frac{\mu_0 N I}{2r}, where μ0\mu_0 is the permeability of free space, NN is the number of turns, II the current, and rr the coil radius. This formula assumes the coil is tightly wound and the magnetic field contributions from each loop add up perfectly.
Moving away from the center along the coil’s axis, the magnetic field weakens because the distance increases, and the magnetic contributions from each loop spread out. The magnetic field on the axis at a distance xx from the center is given by B=μ0NIr22(r2+x2)3/2B = \frac{\mu_0 N I r^2}{2(r^2 + x^2)^{3/2}}
This expression includes the geometric factor (r2+x2)3/2(r^2 + x^2)^{3/2} in the denominator, showing how the field diminishes with distance xx. In this problem, the field at the center is about 12.14 millitesla (mT), but at 7.20 cm away—more than four times the radius—it drops sharply to approximately 0.103 mT, roughly 100 times weaker.
This sharp drop in field strength illustrates the localized nature of magnetic fields from coils and why magnetic devices often need to place sensors or magnetic materials very close to the coil for effective operation. The calculation shows the importance of coil dimensions and positioning in practical applications like electromagnets, inductors, or magnetic resonance imaging (MRI).