Question

The magnetic field at the centre of a circular coil of \(50\) turns and radius \(10\) cm carrying a current of \(1\) A, in tesla is

• A. \(\pi\times10^{-4}\)
• B. \(\pi\times10^{-3}\)
• C. \(2\pi\times10^{-3}\)
• D. \(\dfrac{\pi}{4}\times10^{-4}\)
• E. \(\dfrac{\pi}{2}\times10^{-4}\)

Hint:

For a circular loop: \[ B=\frac{\mu_0 NI}{2R} \] Always convert radius into metres.

Answer 1

The Correct Option is A

Magnetic field at the centre of a circular coil: \[ B=\frac{\mu_0 NI}{2R} \] Given: \[ N=50,\quad I=1,\quad R=10\text{ cm}=0.1\text{ m} \] \[ B=\frac{4\pi\times10^{-7}\times 50}{2\times 0.1} \] \[ B=\frac{200\pi\times10^{-7}}{0.2} =1000\pi\times10^{-7} =\pi\times10^{-4}\text{ T} \] Hence, \[ \boxed{(A)\ \pi\times10^{-4}} \]