High School

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Two circular coils of radii 12.8 cm and 38.4 cm are connected in series to a battery. Find the ratio of the magnitudes of the magnetic fields at their centers.

Answer :

To find the ratio of the magnitudes of the magnetic fields at the centers of two circular coils, we can use the formula for the magnetic field at the center of a circular loop:

[tex]B = \frac{\mu_0 \cdot I \cdot N}{2 \cdot R}[/tex]

Where:

  • [tex]B[/tex] is the magnetic field at the center of the coil.
  • [tex]\mu_0[/tex] is the permeability of free space. This is a constant [tex]4\pi \times 10^{-7}[/tex] T·m/A.
  • [tex]I[/tex] is the current flowing through the coil.
  • [tex]N[/tex] is the number of turns in the coil. Since both coils are connected in series, and typically considered to have the same number of turns, we'll assume it is the same for simplicity.
  • [tex]R[/tex] is the radius of the coil.

In this problem, we want the ratio of the magnetic fields, not their absolute values. If we assume the current [tex]I[/tex] and the number of turns [tex]N[/tex] are the same for both coils and connected in series to a battery, they cancel out in the ratio.

Given the radii of the coils are:

  1. [tex]R_1[/tex] = 12.8 cm = 0.128 m
  2. [tex]R_2[/tex] = 38.4 cm = 0.384 m

The ratio of the magnetic field magnitudes [tex]B_1[/tex] to [tex]B_2[/tex] is given by:

[tex]\frac{B_1}{B_2} = \frac{\frac{\mu_0 I N}{2R_1}}{\frac{\mu_0 I N}{2R_2}} = \frac{R_2}{R_1}[/tex]

Substituting the values of the radii:

[tex]\frac{B_1}{B_2} = \frac{0.384}{0.128} = 3[/tex]

Thus, the ratio of the magnitudes of the magnetic fields at their centers is 3:1.

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