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
R^{2}

= 4 x 3.14 x 10^{-7} x 49 x 0.015 / 2 x 3.14 x
0.02^{2}

= 4 x 10^{-7} x 49 x 0.015 / 2 x 0.02^{2}

= 2 x 10^{-7} x 49 x 0.015 / 0.02^{2}

= 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**

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