Question

what current must flow in a circular coil of 20 loops of radius 40 cm to produce a magnetic field of 3 pi into 10 ^ - 2T at its centre​

Answer 1

Given :

Radius of the circular coil ( R ) = 40 cm .

Convert it into metres .

R = 40 cm ⇒ 40/100 m ⇒ 0.4 m .

Number of loops ( n ) = 20 .

Magnitude of magnetic field ( B ) = 3 π × 10⁻² T .

We know that by Biot Savart's Law we have :

$B=\dfrac{\mu_0\times 2 \pi nI}{4\pi\times r}\\\\\implies 3\pi \times 10^{-2}T=\dfrac{4\times 10^{-7}\pi\times 2\pi\times 20\times I}{4\pi \times 0.4}A\\\\\implies I=\dfrac{3\pi \times 10^{-2}\times 4\pi \times 0.4}{4\times 10^{-7}\pi \times 40\pi}A\\\\\implies I=\dfrac{3\times 10^7\times 4\times 0.4}{4\times 40\times 10^2}A\\\\\implies I =3\times 10^{7-2}\times \dfrac{1}{100}A\\\\\implies I=3\times 10^{5-2}A\\\\\implies I=3\times 10^3A\\\\\implies I=3.0\times 10^3A$

NOTE :

Final Answer should be in scientific notation .

Do not forget to change cm to m .

The unit of current will be in Ampere .

We can also write the answer as 3 kilo ampere .

But writing in scientific notation is safe as per the rules of examination .