Question
A current carrying toroid winding is internally filled with lithium having susceptibility $x = 2.1 \times 10^{−5}.$ What is the percentage increase in the magnetic field in the presence of lithium over that without it?
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Answer
To determine the percentage increase in the magnetic field inside a toroid when it is filled with lithium $($which is a diamagnetic material given its positive susceptibility$),$ we first need to understand how the magnetic susceptibility $(x)$ affects the magnetic field inside materials.
The magnetic susceptibility, $x,$ is a dimensionless proportionality constant that indicates the degree to which a material can be magnetized in response to an external magnetic field. The total magnetic field $(B)$ inside a material in the presence of an external magnetic field $(B_0)$ can be expressed as:
$B=\mu B_0$
where $\mu$ is the permeability of the material, and it is related to the magnetic susceptibility $x$ by the relation:
$\mu=\mu_0(1+x)$
Here, $\mu_0$ is the permeability of free space, and its value is $4 \mu \times$$10^{-7} H / m \text {. }$
Given the magnetic susceptibility of lithium, $x=2.1 \times 10^{-5}$, the percentage increase in the magnetic field due to the presence of lithium over that without it $($i.e., in free space or air, where $x=0$, hence $\mu=\mu_0 )$ can be calculated as follows:
Calculate the permeability of lithium, $\mu_{ Li }$ :
$\mu_{Li}=\mu_0(1+x)$
$\mu_{Li}=\left(4 \pi \times 10^{-7}\right)\left(1+2.1 \times 10^{-5}\right)$
Calculate the magnetic field inside lithium, $B _{ Li }$ using an arbitrary $B _0$, since we are interested in the percentage increase:
Percentage Increase $=\left(\frac{B_{L i}-B_0}{B_0}\right) \times 100 \%$
Percentage Increase $=\left(\frac{\mu_{L i}-\mu_0}{\mu_0}\right) \times 100 \%$
Percentage Increase $=$
$\left(\frac{\left(4 \pi \times 10^{-7}\right)\left(1+2.1 \times 10^{-5}\right)-\left(4 \pi \times 10^{-7}\right)}{4 \pi \times 10^{-7}}\right) \times 100 \%$
Percentage Increase $=\left(1+2.1 \times 10^{-5}-1\right) \times 100 \%$
Percentage Increase $=2.1 \times 10^{-5} \times 100 \%$
Percentage Increase $=2.1 \times 10^{-3} \%$
Thus, the percentage increase in the magnetic field inside a toroid when filled with lithium over that without it is approximately $0.0021\%.$
The magnetic susceptibility, $x,$ is a dimensionless proportionality constant that indicates the degree to which a material can be magnetized in response to an external magnetic field. The total magnetic field $(B)$ inside a material in the presence of an external magnetic field $(B_0)$ can be expressed as:
$B=\mu B_0$
where $\mu$ is the permeability of the material, and it is related to the magnetic susceptibility $x$ by the relation:
$\mu=\mu_0(1+x)$
Here, $\mu_0$ is the permeability of free space, and its value is $4 \mu \times$$10^{-7} H / m \text {. }$
Given the magnetic susceptibility of lithium, $x=2.1 \times 10^{-5}$, the percentage increase in the magnetic field due to the presence of lithium over that without it $($i.e., in free space or air, where $x=0$, hence $\mu=\mu_0 )$ can be calculated as follows:
Calculate the permeability of lithium, $\mu_{ Li }$ :
$\mu_{Li}=\mu_0(1+x)$
$\mu_{Li}=\left(4 \pi \times 10^{-7}\right)\left(1+2.1 \times 10^{-5}\right)$
Calculate the magnetic field inside lithium, $B _{ Li }$ using an arbitrary $B _0$, since we are interested in the percentage increase:
Percentage Increase $=\left(\frac{B_{L i}-B_0}{B_0}\right) \times 100 \%$
Percentage Increase $=\left(\frac{\mu_{L i}-\mu_0}{\mu_0}\right) \times 100 \%$
Percentage Increase $=$
$\left(\frac{\left(4 \pi \times 10^{-7}\right)\left(1+2.1 \times 10^{-5}\right)-\left(4 \pi \times 10^{-7}\right)}{4 \pi \times 10^{-7}}\right) \times 100 \%$
Percentage Increase $=\left(1+2.1 \times 10^{-5}-1\right) \times 100 \%$
Percentage Increase $=2.1 \times 10^{-5} \times 100 \%$
Percentage Increase $=2.1 \times 10^{-3} \%$
Thus, the percentage increase in the magnetic field inside a toroid when filled with lithium over that without it is approximately $0.0021\%.$
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