Question

calculate the magnetic flux density of the magnetic field at the centre of a circular coilvof 100 turns,carrying current 2a and having a radius of 0.2m

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Answer 1

$\sf{\huge{\underline{\orange{Answer :-}}}}$

• The centre of a circular coil is of 100 turns.

• It carries a current of 2A.

• It have radius of 0.2m.

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• The magnetic flux density of the magnetic field .

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We know ,

• n = 100

• I = 2A

• R = 0.2m

We know, Biot savart law states

➝ B = μonI / 2R

➝ B = 4π × ${10}^{ - 7}$ × 100 × 2 / 2 × 0.5

➝ B = 4π × ${10}^{ - 7}$ × 50

➝ B = 200 × ${10}^{ - 7}$

➝ B = 2 × ${10}^{ - 5}$

Hence, The magnetic flux density of the magnetic field is 2 × ${10}^{ - 5}$ .

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Answer 2

Given :-

• The centre of a circular coil is of 100 turns.
• It carries a current of 2A.
• It have radius of 0.2m.

To find :-

• The magnetic flux density of the magnetic field.

Solution :-

★ As we know that,

• n = 100
• I = 2A
• R = 0.2m

★ As we know that,

$~~~~~~~~~~$ Biot savart law states,

$\implies$ B = ${\tt{\frac{μonl}{2R}}}$

$\implies$ B = ${\tt{4π \times {10}^{ - 7} \times 100 \times \frac{2}{2} \times 0.5}}$

$\implies$ B = ${\tt{4π \times {10}^{ - 7} \times 50}}$

$\implies$ B = ${\tt{200 \times {10 }^{ - 7} }}$

$\implies$ B = ${\tt{2 \times {10}^{ - 5} }}$

Hence,

• The magnetic flux density of the magnetic fields is ${\tt\underline{2 \times {10}^{ - 5} }}$