Question

A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 a



a.what is the magnetic field at the centre?



b.what is the magnetic moment of the coil?

Answer 1

a. by using the formula of magnetic field for circle answer is 2.0096*10^-9

Answer 2

Answer:

Given data states that the number of turn as n = 100 and the current I is 3.2 A with radius of the coil being r = 10 cm which is 0.1 m. therefore to find the magnetic field at centre of the coil, we simply need to put the given values in to Magnetic field formula, which is

                               $B = \mu_o n i / 2 r\\=> B = ( 4 p \times 10 ^{-7} \times 100 \times 3.2) / 2 \times 0.1\\=> B = 2 \times 10 ^{-3} T.$

And similarly for the Magnetic moment of the coil, simply input the answer into

                                $M = n i A \\=> M = n i \times p r ^2\\ => M = 100 \times 3.2 \times 3.14 \times 0.1 \times 0.1\\=> M = 10 A m^2$