Question

a circular loop of radius 3 cm is having a current of 12.5 ampere the magnitude of the magnetic field at a distance of 4 cm on its axis is​

Answer 1

Answer:5.65×10^-5 T

Explanation:

Answer 2

Answer:

here,

I = 12.5 A

we know,

$i = \frac{q}{t} \\ > 12.5 = \frac{q}{t} \\ > q = 12.5t$

again,

distance, r=4 cm=0.04 m

charge, q= 12.5t

so,

electric field,

$= k \frac{q}{r^{2} } \\ = k \frac{12.5t}{(0.04) ^{2} } \\ = 9 \times {10}^{9} \times \frac{12.5}{0.04 ^{2} } t \\ = 9 \times {10}^{9} \times 0.02t \\ = 0.18 \times {10}^{9} \\ = 1.8 \times {10}^{8}$

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