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Question: The inner and outer radius of a toroid core are \(28\,cm\) and \(29\,cm\) respectively and around th...

The inner and outer radius of a toroid core are 28 cm28\,cm and 29 cm29\,cm respectively and around the core 3700 turns of a wire are wound. If the current wire is 10 A10\,A, then the magnetic field inside the core of the toroid is:
A. 2.60×10−2T2.60 \times {10^{ - 2}}T
B. 2.60×10−3T2.60 \times {10^{ - 3}}T
C. 4.52×10−2T4.52 \times {10^{ - 2}}T
D. 4.52×10−3T4.52 \times {10^{ - 3}}T

Explanation

Solution

We’ve been given a toroid whose inner radius is 28 cm and the outer radius is 29 cm, from this we can find the radius of the amperian loop of the toroid. The magnetic field inside the toroid will be given by the ampere’s circuital law, which states that the magnetic field along an imaginary closed path (i.e. amperian loop) is equal to the product of current enclosed by the amperian loop and the permeability of the medium.

Formula Used:
B = μ0N I2πR{\text{B = }}\dfrac{{{\mu _0}{\text{N I}}}}{{2\pi R}}
Were,μ0\mu _0 is the permeability given by 4π×10−7N/A24\pi \times 10^{-7} N/A^2, NN is the number of turns, II is the current through the coil and RR is the radius of the amperian loop.

Complete step by step answer:
In the problem, they’ve given a toroid of inner radius 28cm and outer radius 29cm. But we will consider the radius of the dotted circle (amperian loop) as the field exists in the core part of the toroid. The radius RR will be,
R=28+292R = \dfrac{{28 + 29}}{2}
⇒R=28.5 cm\Rightarrow R = 28.5{\text{ cm}}

From Ampere’s circuital law, we have that the relationship between current and the magnetic field as,
B = μ0NI2πR{\text{B = }}\dfrac{{{\mu _0}NI}}{{2\pi R}}
We found the radius of the amperian loop to be 28.5 cm which can be written as 28.5×10−2 m.{\text{28}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 2}}{\text{ m}}{\text{.}}
They’ve given that the number of turns is 3700 and the current passing through the wire is 10 A. We also have the value of permeability constant.

Substituting these values given in the problem we have
B = μ0NI2πR{\text{B = }}\dfrac{{{\mu _0}NI}}{{2\pi R}}
⇒B=4π×10−7×3700×102π×28.5×10−2\Rightarrow B = \dfrac{{4\pi \times {{10}^{ - 7}} \times 3700 \times 10}}{{2\pi \times 28.5 \times {{10}^{ - 2}}}}
∴B=2.60×10−2 T\therefore B = 2.60 \times {10^{ - 2}}\,T
Therefore, the magnetic field inside the toroid is 2.60×10−2T2.60 \times {10^{ - 2}}T.

Hence, the correct option is (A).

Note: It should be remembered that the amperian loop is an imaginary loop, used for measuring the magnetic field in a conductor. It’s important to understand that there exists a magnetic field outside the toroid. But we assume it to be zero as they cancel each other out. This can easily be understood by the right-hand thumb rule. If you point the thumb in the direction of the current the curled fingers will give the direction of the magnetic field due to the current. And it is mentioned that the core of toroid is non-ferromagnetic. Thus, we can take the permeability of free space. But if it was a soft iron core, we’ll have to consider the relative permeability.