Question 23: 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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