Question 25: Magnetism and Matter

Question

The free space inside a current carrying toroid is filled with a material of susceptibility $2 	imes 10^{-2}$. The percentage increase in the value of magnetic field inside the toroid will be

Accepted Answer

The magnetic susceptibility ($\chi_m$) of a material is the measure of how much the material becomes magnetized in response to an external magnetic field. When the material is placed in a magnetic field, the net magnetic field ($B$) inside the material is the sum of the external magnetic field ($B_0$) and the field produced by the material itself ($B_{material}$). This relationship can be expressed as :
$B = B_0 + B_{material}$
Magnetic susceptibility is related to the relative permeability ($\mu_r$) of the material :
$\mu_r = 1 + \chi_m$
The magnetic field inside the toroid can be calculated using the following formula :
$B = \mu_0 \mu_r H$
where $\mu_0$ is the permeability of free space, $\mu_r$ is the relative permeability of the material, and $H$ is the magnetic field strength.
Now, let's consider the percentage increase in the magnetic field when the material is placed inside the toroid. The initial magnetic field ($B_0$) is given by :
$B_0 = \mu_0 H$
After the material is placed inside the toroid, the magnetic field becomes :
$B = \mu_0 \mu_r H = \mu_0 (1 + \chi_m) H$
The percentage increase in the magnetic field can be calculated as :
$\frac{B - B_0}{B_0} 	imes 100\% = \frac{\mu_0 (1 + \chi_m) H - \mu_0 H}{\mu_0 H} 	imes 100\%$
Substitute the value of $\chi_m = 2 	imes 10^{-2}$ :
$\frac{B - B_0}{B_0} 	imes 100\% = \frac{(1 + 2 	imes 10^{-2}) - 1}{1} 	imes 100\% = 2\%$
Thus, the percentage increase in the value of the magnetic field inside the toroid is 2%.

Answer

0.1%

Answer

1%

Answer

2%

Answer

0.2%

Magnetism and Matter

The free space inside a current carrying toroid is filled with a material of susceptibility 2 \times 10^{-2}. The percentage increase in the value of magnetic field inside the toroid will be

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