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Chapter 26: Problem 58

Your company is developing a device incorporating a \(20-\mathrm{cm}-\) diameter coil carrying 0.50 A that, when properly oriented, will just cancel Earth's \(50-\mu\) T magnetic field at the coil's center. How much wire must you requisition for each coil?

Short Answer

Expert verified
You need to requisition 1000 meters of wire for each coil.

Step by step solution

01

Identify Knowns

The known parameters are Magnetic Field (B) \(= 50\mu T = 50 \times 10^{-6} T\), Current (I) \(= 0.50 A\) and Diameter of the coil \(= 20cm = 0.2m\). Radius (R) of the coil will be diameter/2 \(= 0.1m\). Permeability of free space \(\mu_0 = 4\pi \times 10^{-7} Tm/A\). We need to find the number of turns in the coil (n).
02

Calculate Number of Turns

Rearrange the formula B = μ₀.n.I / (2.R), for number of turns (n), we get \(n = B \times 2 \times R / (\mu_{0} \times I)\). Substituting the known values, \(n = 50 \times 10^{-6} \times 2 \times 0.1 / (4\pi \times 10^{-7} \times 0.50) = 1592\) turns.
03

Calculate Wire Requirement

The amount of wire can be calculated as the circumference of the coil (which is \(2\pi R\)) times the number of turns (n). Therefore, required wire \(= n \times 2\pi R = 1592 \times 2\pi \times 0.1 = 1000m\). So, 1000 meters of wire should be requisitioned for each coil.

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