Question

toroid of 4000 turns has outer radia
ad inney radius of 25cm. In the carrer
10 A . calculate the magnetic field of​

Answer 1

GiveN :-

• A toroid of 4000 turns has a outer radius of 25cm .
• The inner radius of the toroid is 25cm .
• A current of 10A is flowing .

To FinD :-

• The magnetic field of toroid .

SolutioN :-

We need to find the Magnetic field of toroid . The given data here is ,

$\red{\frak{Given}}\begin{cases} \textsf{ Outer radius of toroid is \textbf{26 cm}.} \\ \textsf{ Inner radius of toroid is \textbf{25 cm}.} \\ \textsf{ Current flowing is \textbf{10A}.} \end{cases}$

We know that the Magnetic field inside the core of toroid can be found by ,

$\sf :\implies \pink{ B =\mu_0 nl}\\\\\sf:\implies B = \mu_0 \bigg( \dfrac{n_{(turns)}}{Circumference} \times I \bigg) \\\\\sf:\implies B = \mu_0 \bigg( \dfrac{4000}{2\pi \times \frac{25+26}{2}}\bigg) \times 11 A \\\\\sf:\implies B = \mu_0 \bigg( \dfrac{4000}{25.5 cm \times 2\pi }\bigg) \times 11A \\\\\sf:\implies B =\mu_0 \dfrac{4000}{ 2\pi \times 25.5 \times 10^{-2} m }\times 11A \\\\\sf:\implies B = 4\pi \times 10^{-7} \times \dfrac{4000}{51\pi \times 10^{-2} m }\times 11 A \\\\\sf:\implies B = 4\times 10^{-5} \times \dfrac{4000}{51}\times 11 \\\\\sf:\implies \underset{\blue{\sf Required \ Magnetic \ field }}{\underbrace{\boxed{\pink{\frak{ B = 3.4 \times 10^{-2} \ Tesla }}}}}$