The magnetic field around the head has been measured to be approximately 3.00×10−8 gauss . Although the currents that cause this field are quite complicated, we can get a rough estimate of their size by modeling them as a single circular current loop 16.0 cm (the width of a typical head) in diameter. What is the current needed to produce such a field at the center of the loop?

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Vespertilio Vespertilio · Answer 1 · 2019-11-03T06:18:01+01:00

Answer:

1.2 micro Ampere

Explanation:

Magnetic field, B = 3 x 10^-8 Gauss = 3 x 10^-12 T

Diameter, d = 16 cm

Radius, r = 16 / 2 = 8 cm = 0.08 m

Let the current is i.

The formula for the magnetic field cue to current carrying coil is goven by

[tex]B=\frac{\mu _{0}}{4\pi }\times \frac{2i}{r}[/tex]

By substituting the values

[tex]3\times 10^{-12}=10^{-7}\times \frac{2i}{0.08}[/tex]

i = 1.2 x 10^-6 A

Thus, the current in the coil is 1.2 micro Ampere.

sp10925 sp10925 · Answer 2 · 2022-06-06T14:25:51+02:00

The current needed to produce such a field at the center of the loop is 1.2 micro Ampere.

The current is the stream of charges which flow inside the conductors when connected across the end of voltage.

Given is the Magnetic field, B = 3 x 10⁻⁸ Gauss = 3 x 10⁻¹²T, Diameter d = 16 cm, then

Radius, r = 16 / 2 = 8 cm = 0.08 m

If the current is i, then the magnetic field due to current carrying coil is given by

B = ( μ₀/4π) x (2i/r)

By substituting the values, we get

i = 1.2 x 10⁻⁶ Amperes

Thus, the current in the coil is 1.2 micro Ampere.

Learn more about current.

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