Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A coil wrapped around toroid has inner radius of $20 \mathrm{~cm}$ and an outer radius of $25 \mathrm{~cm}$. If the wire wrapping makes 800 turns and carries a current of $12 \mathrm{~A}$. The maximum and minimum values of the magnetic field with in the toroid are
PhysicsMagnetic Effects of CurrentAP EAMCETAP EAMCET 2021 (25 Aug Shift 1)
Options:
  • A $9.6 \mathrm{mT}, 7.68 \mathrm{mT}$
  • B $4 \mathrm{mT}, 2.5 \mathrm{mT}$
  • C $7 \mathrm{mT}, 5.6 \mathrm{mT}$
  • D $6.6 \mathrm{mT}, 3.3 \mathrm{mT}$
Solution:
1449 Upvotes Verified Answer
The correct answer is: $9.6 \mathrm{mT}, 7.68 \mathrm{mT}$
Given, inner radius, $r_1=20 \mathrm{~cm}=0.2 \mathrm{~m}$
Outer radius, $r_2=25 \mathrm{~cm}=0.25 \mathrm{~m}$
Number of turns, $N=800$
Current in the toroid, $I=12 \mathrm{~A}$
Magnetic field due to $N$ circular rings (toroid),
$$
B=\frac{\mu \cdot N I}{2 \pi r}
$$
Magnetic field is maximum at $r_{\min }=0.2 \mathrm{~m}$
$$
\begin{aligned}
B_{\max } & =\frac{4 \pi \times 10^{-7} \times 800 \times 12}{2 \pi(0.2)}=9.6 \times 10^{-3} \mathrm{~T} \\
& =9.6 \mathrm{mT}
\end{aligned}
$$
Magnetic field is minimum at $r_{\max }=0.25 \mathrm{~m}$
$$
\begin{aligned}
B_{\min } & =\frac{4 \pi \times 10^{-7} \times 800 \times 12}{2 \pi \times(0.25)}=7.68 \times 10^{-3} \mathrm{~T} \\
& =7.68 \mathrm{mT}
\end{aligned}
$$
So, the maximum value of magnetic field within the toroid is $9.6 \mathrm{mT}$ and minimum value of magnetic field within the toroid is $7.68 \mathrm{mT}$.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.