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 r-soy February 18th, 2011 10:21 AM

Compare the magnetic field strength inside & outside the so

1 - Compare the magnetic field strength inside and outside the solenoid ?
2 - A magnetic field inside a solenoid is 0.2 T. what is the number of turns/cm of the solenoid if the solenoid carries a current of 10 A ?
3 - The magnetic field a distance of 4 cm from a long straight current - carrying wire is 2 X 10 ^ -5 T . what is the current in the wire ? what is the magnetic field at distance of 12 cm from the wire ?

 MarkFL February 18th, 2011 11:59 AM

Re: Compare the magnetic field strength inside & outside the

1.) If we assume the case of an ideal solenoid, i.e., the turns are closely spaced and the length is long compared to the radius, then we can assume the magnetic field inside is uniform and the field outside is zero. The uniform field strength inside will be given by:

$B=\mu_0nI$ where:

$\mu_0=4\pi\times10^{-7}\text{ T\cdot m/A}$ is a constant called the permeability of free space
$n=\frac{N}{l}$ is the number of turns per unit length
$I$ is the current

2.) Using the expression above, we find:

$0.2\text{ T}=$$4\pi\times10^{-7}\text{ T\cdot m/A}$$n$$10\text{ A}$$$

$n=\frac{2}{4\pi\times10^{-5}\text{ m}}=\frac{10^5}{2\pi}\:\text{\frac{turns}{m}}\cdot \frac{1\text{ m}}{100\text{ cm}}=\frac{1000}{\2\pi}\:\text{\frac{turns}{cm}}$

3.) The magnetic field strength outside a long straight current-carrying wire is given by:

$B=\frac{\mu_0I}{2\pi r}$ where r is the distance from the center of the wire's cross section. Using the data given, we have:

$2\times10^{-5}\text{ T}=\frac{$$4\pi\times10^{-7}\text{ T\cdot m/A}$$I}{2\pi$$0.04\text{ m}$$}$

$I=\frac{$$2\times10^{-5}$$$$8\pi\times10^{-2}$$}{4\pi\times10^{-7}}\:\text{ A}=4\text{ A}$

To find the field strength at 12 cm, we may use the fact that the field strength varies inversely as r, so if we triple the distance, the field strength is 1/3 the original.

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