Question

A long cylindrical wire (radius = 2.0 cm) carries a current of 49 A that is uniformly distributed over the cross-section of the wire. What is the magnitude of the magnetic field at a point which is 1.5 cm from the axis of the wire?

0.18 mT

0.44 mT

0.37 mT

0.49 mT

Answer 1

given

radius = 2.0 cm = R

R = 0.02 m

current of 49 A = i

r = the magnitude of the magnetic field at a point which is 1.5 cm

r = 0.015 m

using equation

B = _o i r / 2 R²

= 4 x 3.14 x 10^-7 x 49 x 0.015 / 2 x 3.14 x 0.02²

= 4 x 10^-7 x 49 x 0.015 / 2 x 0.02²

= 2 x 10^-7 x 49 x 0.015 / 0.02²

= 0.0003675 T

= 0.00037 T

B = 0.37 x 10^-3 T

B = 0.37 mT

the magnitude of the magnetic field at a point which is 1.5 cm from the axis of the wire is B = 0.37 mT