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2 years ago

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Consider a long, closely wound solenoid with 10,000 turns per meter. What current, in amperes, is needed in the solenoid to prod

uce a magnetic field inside the solenoid, near its center, that is 104 times the Earth’s magnetic field of 4.95 × 10-5 T?

1 answer:

Makovka662 [10]2 years ago

4 0

Answer:

0.4344A

Explanation:

From Ampere's law, it can be shown that the magnetic field B inside a long solenoid is

$B= \mu_0NI$

Where

B= Magnetic field strenght at distance d

I= current

$\mu_0 =$ Permeability of free space ( $4\pi*10^{-7} Tm/A$ )

N= Number of loops

Our values are defined as follow,

$N=10000$

$B=5.25*10^{-5}$ T

$B'=5.25*10^{-5} * 104 = 5.46*10^{-3}T$

As a current required to become 104 times the Earth's magnetic field is required, we use B '

$B'= \mu_0NI$

$5.46*10^{-3}=4\pi*10^{-7}*10000*I$

$I=\frac{5.46*10^{-3}}{4\pi*10^{-7}*10000}$

$I=0.4344A$

<em>Therefore is needed 0.4344A in the solenoid to produce a magnetic field inside the solenoid, near its center, that is 104 times the Earth's magnetic field.</em>

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