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Arlecino [84]
3 years ago
15

A closely wound circular coil has a radius of 6.00 cmand carries a current of 2.65 A. How many turns must it have if the magneti

c field at its center is 6.31Ã10^â4T?
Physics
1 answer:
ad-work [718]3 years ago
8 0

Answer:

Given:

radius of the coil, R = 6 cm = 0.06 m

current in the coil, I = 2.65 A

Magnetic field at the center, B = 6.31\times 10^{4} T

Solution:

To find the number of turns, N, we use the given formula:

B = \frac{\mu_{o}NI}{2R}

Therefore,

N = \frac{2BR}{\mu_{o}I}

N = \frac{2\times 6.31\times 10^{4}\times 0.06}{4\pi \times 10^{- 7}\times 2.65}

N = 22.74 = 23 turns (approx)

 

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