Question

Calculate the magnetic flux density of a circular coil of 50 turns having radius of 0.5 m and carrying a current of 5A​

Answer 1

Answer:

MAGNETIC FLUX DENSITY ( B ) = 250 μ° Weber / m²

Explanation:

★ MAGNETIC FLUX DENSITY ★

★ GIVEN ;

» Circular Coil of turns ( N ) = 50

» Radius of Coil ( R ) = 0.5 m

» Current in Loop ( i ) = 5 A

» Permeability Constant = μ°

★ Magnetic Flux of Loop ( B ) = ??

_______ [ BY USING FORMULA ] ______

★ [ B = ( μ° × i ) × N / 2 R ] ★

★ B = [ μ° × 5 ] × 50 / 2 × 0.5

» B = [ 5 × μ° ] × 50 / 1

» B = [ 250 μ° ]

★ MAGNETIC FLUX DENSITY ( B ) = 250 μ° Tesla [ WEBER / m² ] ★

________________________________________

ANSWER :- HENCE ; MAGNETIC FLUX DENSITY OF CURRENT CARRYING LOOP IS ;

➡ B = 250 μ° T ⬅

Answer 2

$\bold {\huge {Your ~answer :-}}$

$\bf\huge\boxed{ 250\:u_0\:Tesla }$

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EXPLAINATION ➣

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$B = \frac{u_0 \:. i}{2r} \times N$

Magnetic flux for a circular loop

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

Here ,

→B is magnetic flux density of circular loop = ?

→i is current flowing through the loop = 5 A

→r is radius of the loop = 0.5 m

→u0 is permeability constant

→N is no of turns = 50

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$B = \frac{u_0 \:. i}{2r} \times N$

Now by putting in formula

$= > B = \frac{(u_0 \times 5) \times 50}{2 \times 0.5}$

⇒B = (5×50)u° / (0.5×2)

⇨ B = 250 u° / 1

⇒B = 250 u° Tesla

Hence, the magnetic flux density is 250 u° Tesla

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