Question

If a wire of length 196 cm carrying a current of 0.98 A is bent in the form of a circular loop of two turns, then the magnetic field at the centre of the loop is approximately:

• A. 8 $\mu$T
• B. 4 $\mu$T
• C. 3 $\mu$T
• D. 6 $\mu$T

Hint:

Magnetic field increases with the number of turns ($N$) and current ($I$).

Answer 1

The Correct Option is A

Step 1: Set up the wire.
A wire of length $196\ cm = 1.96\ m$ is bent into a circular loop of two turns. A current of $0.98\ A$ flows through it. We want the magnetic field at the centre.

Step 2: Recall the field at a coil centre.
For a coil with $N$ turns and radius $r$, the field at the centre is $B = \dfrac{\mu_0 N I}{2r}$. We need to find $r$ first.

Step 3: Find the radius from the wire length.
The whole wire forms $N = 2$ turns, so its length equals the total of two circumferences: $L = N\,(2\pi r) = 4\pi r$.

Step 4: Solve for r.
So $r = \dfrac{L}{4\pi} = \dfrac{1.96}{4\pi} \approx 0.156\ m$.

Step 5: Put the numbers into the field formula.
Using $\mu_0 = 4\pi\times10^{-7}$, $N = 2$, $I = 0.98\ A$ and $r \approx 0.156\ m$: \[ B = \frac{(4\pi\times10^{-7})(2)(0.98)}{2(0.156)} \]

Step 6: Compute the value.
Working this out gives about $8\times10^{-6}\ T$, which is $8\ \mu T$. \[ \boxed{8~\mu T} \]