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Figure 29-67 shows a cross section across a diameter of a long cylindrical conductor of radius a=2.00cmcarrying uniform current 170A. What is the magnitude of the current’s magnetic field at radial distance (a) 0 (b) 1.00 cm, (c) 2.00 cm (wire’s surface), and (d) 4.00 cm?

Short Answer

Expert verified

(a) The magnitude of the current’s magnetic field at a radial distance 0 cm is zero.

(b)The magnitude of the current’s magnetic field at a radial distance 1.00cm is8.50×10-4T

(c) The magnitude of the current’s magnetic field at a radial distance 2.00 cm is1.70×10-3T

(d) The magnitude of the current’s magnetic field at a radial distance 4.00 cm is8.50×10-4T

Step by step solution

01

Listing the given quantities

  • The radius of the conductor isa=2.00cm=0.0200m
  • Current through the conductor isi=170A
02

Understanding the concept of the magnetic field at a point

Ampere’s law states that,

B·ds=μ0i

The line integral in this equation is evaluated around a closed-loop called an Amperian loop.The current ion the right side is the net current encircled by the loop.

Using Ampere’s law, we can derive the magnetic fieldinside the wire(r<a)as,

B=(μ0i2πa2)r (i)

Here,iis the current,ais area of cross-section of the wire,ris the radius.

For the first two conditions r<a, therefore, using, we can find the magnitude of the current’s magnetic field at the corresponding radii. We can also use the same equation for the third condition a=r. For the fourth condition,r>a, therefore, we can use equation (i) to find the corresponding magnitude of the current’s magnetic field.

The magnetic field outside the wire (r>a)is given by,

B=μ0i2πr (ii)

03

(a) Calculations of the magnitude of the current’s magnetic field at a radial distance 0 cm

Forr=0,

Sincea>r, therefore, the magnetic field is given by,

B=μ0i2πa2r=μ0i2πa2×0=0

The magnitude of the current’s magnetic field at a radial distance 0 cm is zero.

04

(b) Calculations of the magnitude of the current’s magnetic field at a radial distance 1.00 cm

Forr=0.0100m

Since,a>r, therefore,fromequation (i), the magnetic field is given by,

B=μ0i2πa2r=4π×10-7T.m/A170A2π0.0200m20.0100m=8.50×10-4T

The magnitude of the current’s magnetic field at a radial distance 1.00 cm is 8.50×10-4T.

05

(c) Calculations of the magnitude of the current’s magnetic field at a radial distance 2.00 cm

Forr=0.0200m

Since,a=r, therefore,fromequation (ii), the magnetic field is given by,

B=μ0i2πa2r=4π×10-7T.m/A170A2π0.0200m20.0200m=1.70×10-3T

The magnitude of the current’s magnetic field at a radial distance 2.00 cm is1.70×10-3T.

06

(d) Calculations of the magnitude of the current’s magnetic field at a radial distance 4.00 cm

Forr=0.0400m,

Since,a<r, therefore,fromequation (ii), the magnetic field is given by

B=μ0i2πr=4π×10-7T.m/A170A2π0.0400m=8.50×10-4T

The magnitude of the current’s magnetic field at a radial distance 4.00 cm is8.50×10-4T.

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Most popular questions from this chapter

In Figure 29-46 two concentric circular loops of wire carrying current in the same direction lie in the same plane. Loop 1 has radius1.50cm and carries 4.00mA. Loop 2 has radius2.50cmand carries 6.00mA.Loop 2 is to be rotated about a diameter while the net magnetic field Bset up by the two loops at their common center is measured. Through what angle must loop 2 be rotated so that the magnitude of that net field is 100nT?

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