A10-gaugebare copper wire ( 2.6mmin diameter) can carry a current of 50Awithout overheating. For this current, what is the magnitude of the magnetic field at the surface of the wire?

Short Answer

Expert verified

Magnitude of the net magnetic field at the surface of wire is 7.7×10-3T.

Step by step solution

01

Identification of the given data

  1. The current is, i=50A.
  2. The diameter is,d=2.6mm.
02

Understanding the concept

We use the formula of magnetic field due to a long straight wire at perpendicular distancefrom the wire to calculate the magnitude of magnetic field.

Formula:

B=μ0i2πd

03

Calculate the magnitude of the net magnetic field at the surface of the wire

The magnetic field due to the long straight wire at any distance from the wire is given by

B=μ0i2πd

We have to find the magnitude of the magnetic field on the surface of the wire whose distance from the center of wire is Rand given by

B=μ0i2πR

Diameter d=2.6mmand radius R=2.62mm=0.0013m.

Substitute all the value in the above equation.

localid="1664177943519" B=4π×10-7TmA×50A2π×0.0013m

localid="1663008809868" B=7.7×10-3T

Hence the net magnetic field at the surface of wire is 7.7×10-3T.

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Most popular questions from this chapter

Figure 29-80 shows a cross-section of a long cylindrical conductor of radius a=4cmcontaining a long cylindrical hole of radiusb=1.50cm. The central axes of the cylinder and hole are parallel and are distanced=2cmapart; currentis uniformly distributed over the tinted area. (a) What is the magnitude of the magnetic field at the center of the hole? (b) Discuss the two special casesb=0andd=0.

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(b) If the projectile starts from the left end of the rails at rest, find the speed vat which it is expelled at the right. Assume that I = 450 kA, w = 12 mm, R = 6.7 cm, L = 4.0 m, and the projectile mass is 10 g.

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