Question

Two toroids with number of turns 400 and 200 have average radii respectively 30 cm and 60 cm. If they carry the same current, the ratio of magnetic fields in these two toroids is:

• A. \( 2:1 \)
• B. \( 1:4 \)
• C. \( 2:3 \)
• D. \( 4:1 \)

Hint:

The magnetic field inside a toroid is inversely proportional to the radius and directly proportional to the number of turns. To find the ratio, use: \[ \frac{B_1}{B_2} = \frac{N_1}{N_2} \times \frac{r_2}{r_1}. \]

Answer 1

The Correct Option is D

Step 1: Using the Formula for Magnetic Field in a Toroid The magnetic field inside a toroid is given by: \[ B = \frac{\mu_0 N I}{2 \pi r}. \] where: - \( N \) is the number of turns,
- \( I \) is the current,
- \( r \) is the average radius of the toroid. Since both toroids carry the same current, the ratio of their magnetic fields is: \[ \frac{B_1}{B_2} = \frac{\left( \frac{\mu_0 N_1 I}{2 \pi r_1} \right)}{\left( \frac{\mu_0 N_2 I}{2 \pi r_2} \right)}. \] Cancelling common terms: \[ \frac{B_1}{B_2} = \frac{N_1}{N_2} \times \frac{r_2}{r_1}. \]
Step 2: Substituting the Given Values - \( N_1 = 400 \), \( N_2 = 200 \),
- \( r_1 = 30 \) cm, \( r_2 = 60 \) cm. \[ \frac{B_1}{B_2} = \frac{400}{200} \times \frac{60}{30}. \] \[ \frac{B_1}{B_2} = 2 \times 2 = 4. \] Thus, the ratio is: \[ \boxed{4:1}. \]