Question

The magnetic field inside a current carrying toroidal solenoid is $0.2\text{ mT}$. What is the magnetic field inside the toroid if the current through it is tripled and the cross-sectional radius is made $\frac{1}{3}^{\text{rd}}$?

• A. $0.2\text{ mT}$
• B. $0.6\text{ mT}$
• C. $0.8\text{ mT}$
• D. $0.9\text{ mT}$

Hint:

Don't fall for trick parameters! In an ideal toroid, the internal magnetic field depends on the major radius of the ring ($R$), but is completely unaffected by changes to the minor cross-sectional radius ($r$) of the tube. Focus only on the current tripling to get $3 \times 0.2 = 0.6\text{ mT}$ instantly.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given an initial magnetic field $B_1 = 0.2\text{ mT}$ inside a toroidal solenoid coil. The operating current is then tripled ($I_2 = 3I_1$) and the tube's cross-sectional radius is reduced to one-third of its original value. We need to determine the new magnetic field intensity, $B_2$.

Step 2: Key Formula or Approach:
The magnetic field induction $B$ inside an ideal toroidal solenoid is given by the formula: $$B = \mu_0 n I = \mu_0 \left(\frac{N}{2\pi R_{\text{toroid}}}\right) I$$ where $n$ is the number of turns per unit length, $N$ is the total number of turns, and $R_{\text{toroid}}$ is the major central loop radius of the toroid ring itself. This formula shows that the internal field depends only on the current $I$ and the total loop circumference, and is independent of the secondary cross-sectional tube radius ($r$).

Step 3: Detailed Explanation:
The formula shows that the internal magnetic field is directly proportional to the current ($B \propto I$) and completely independent of the tube's cross-sectional wire radius $r$. Changing the cross-sectional radius has no effect on the magnetic field. Therefore, we only need to account for the change in current: $$\frac{B_2}{B_1} = \frac{I_2}{I_1}$$ Given that the current is tripled ($I_2 = 3I_1$): $$\frac{B_2}{0.2\text{ mT}} = \frac{3I_1}{I_1} = 3$$ Multiply to solve for $B_2$: $$B_2 = 3 \times 0.2\text{ mT} = 0.6\text{ mT}$$ This matches option (B).

Step 4: Final Answer:
The new magnetic field inside the toroid is $0.6\text{ mT}$, which corresponds to option (B).