(A) Given: Number of turns $N = 200$,radius $r = 20 \ cm = 0.2 \ m$,and current $I = 5 \ A$.The magnetic field $B$ at the centre of a circular coil is

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(A) Given: Number of turns $N = 200$,radius $r = 20 \ cm = 0.2 \ m$,and current $I = 5 \ A$.The magnetic field $B$ at the centre of a circular coil is given by the formula:$B = \frac{\mu_0 NI}{2r}$Substituting the values:$B = \frac{(4\pi \times 10^{-7} \ T \cdot m/A) \times 200 \times 5}{2 \times 0.2 \ m}$$B = \frac{4 \times 3.14159 \times 10^{-7} \times 1000}{0.4}$$B = \frac{12.566 \times 10^{-4}}{0.4}$$B = 31.4159 \times 10^{-4} \ T = 3.14 \times 10^{-3} \ T$.

Class 12 Physics Moving Charges and Magnetism Biot-Savart's Law and its application

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$A$ tightly wound coil of $200$ turns and of radius $20 \ cm$ carries a current of $5 \ A$. The magnetic field at the centre of the coil is:

AP EAMCET 2024 All AP EAMCET

A
$3.14 \times 10^{-3} \ T$
B
$3.14 \times 10^{-2} \ T$
C
$6.28 \times 10^{-4} \ T$
D
$6.28 \times 10^{-3} \ T$

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Moving Charges and Magnetism

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Biot-Savart's Law and its application

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$A$ tightly wound coil of $200$ turns and of radius $20 \ cm$ carries a current of $5 \ A$. The magnetic field at the centre of the coil is:

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