Two toroids 1 and 2 have total no. of turns 200 and 100 respectively with average radii 40 cm and 20 cm respec

Question

Two toroids 1 and 2 have total no. of turns 200 and 100 respectively with average radii 40 cm and 20 cm respectively. If they carry the same current i, what will be the ratio of the magnetic fields along the two loops?

Accepted Answer

1. **Formula:** The magnetic field $B$ inside a toroid is given by the formula:
   $$B = \frac{\mu_0 N I}{2\pi r}$$
   where $N$ is the total number of turns, $I$ is the current, and $r$ is the average radius of the toroid .
2. **Given Values:**
   - **Toroid 1:** $N_1 = 200$, $r_1 = 40 	ext{ cm} = 0.4 	ext{ m}$, Current = $i$.
   - **Toroid 2:** $N_2 = 100$, $r_2 = 20 	ext{ cm} = 0.2 	ext{ m}$, Current = $i$.
3. **Calculation of Ratio ($B_1/B_2$):**
   $$ \frac{B_1}{B_2} = \frac{\frac{\mu_0 N_1 i}{2\pi r_1}}{\frac{\mu_0 N_2 i}{2\pi r_2}} $$
   $$ \frac{B_1}{B_2} = \left( \frac{N_1}{r_1} 
ight) 	imes \left( \frac{r_2}{N_2} 
ight) $$
   $$ \frac{B_1}{B_2} = \left( \frac{200}{40} 
ight) 	imes \left( \frac{20}{100} 
ight) $$
   $$ \frac{B_1}{B_2} = 5 	imes 0.2 = 1 $$
   Thus, the ratio is 1:1.

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