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: \[ \boxed{4:1}. \]

Answer 2

To find the ratio of the magnetic fields in the two toroids, we need to use the formula for the magnetic field inside a toroid, which is:

Magnetic field (B) = \( \dfrac{\mu_0 N I}{2\pi r} \)

where \( \mu_0 \) is the permeability of free space, \( N \) is the number of turns, \( I \) is the current, and \( r \) is the average radius of the toroid.

Given:

• Toroid 1: \( N_1 = 400 \), \( r_1 = 30 \) cm (or 0.3 m)
• Toroid 2: \( N_2 = 200 \), \( r_2 = 60 \) cm (or 0.6 m)
• Current \( I \) is the same for both toroids.

The magnetic fields for the two toroids are:

\( B_1 = \dfrac{\mu_0 \cdot 400 \cdot I}{2\pi \cdot 0.3} \)

\( B_2 = \dfrac{\mu_0 \cdot 200 \cdot I}{2\pi \cdot 0.6} \)

To find the ratio \( \dfrac{B_1}{B_2} \):

\( \dfrac{B_1}{B_2} = \dfrac{\mu_0 \cdot 400 \cdot I}{2\pi \cdot 0.3} \times \dfrac{2\pi \cdot 0.6}{\mu_0 \cdot 200 \cdot I} \)

Simplifying, we have:

\( \dfrac{B_1}{B_2} = \dfrac{400 \cdot 0.6}{200 \cdot 0.3} = \dfrac{240}{60} = 4 \)

The ratio of the magnetic fields is 4:1.