Find the magnetic field produced by a 5A current at the centre of a 50 turn coil of 8 cm diameter.
Given, $\left(\mu_0=4\pi \times 10^{-7}\frac{weber}{amp.m}\right)$
Answer & explanation
Correct answer: option 1
The correct answer is option (1) : $39.2\times 10^{-4} weber/m^2$
Magnetic field at centre of coil $=\frac{\mu_0 \times i }{2R}$
⇒ Magnetic field due to 5 A current at centre of a 50 turn coil of 8 cm diameter is $=\frac{50 \times \mu_0 \times i }{2R}$
$=\frac{50\times 4\pi \times 10^{-7} \times 5 }{2 \times 4 \times 10^{-2}}=125\times \pi \times 10^{-5}$
$= 392.6 \times 10^{-5}$
$=39.2\times 10^{-4} weber/m^2$