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Fundamentals Of Physics
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Short Answer

Figure 29-67 shows a cross section across a diameter of a long cylindrical conductor of radius a= 2.00 cm carrying uniform current 170 A. 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?

(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 is 8.50×10-4T

(c) The magnitude of the current’s magnetic field at a radial distance 2.00 cm is 1.70×10-3T

(d) The magnitude of the current’s magnetic field at a radial distance 4.00 cm is 8.50×10-4T

See the step by step solution

Step by Step Solution

Step 1: Listing the given quantities

  • The radius of the conductor is a=2.00 cm=0.0200 m
  • Current through the conductor is i=170 A

Step 2: 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 field inside the wire (r < a) as,

B=(μ0i2πa2)r (i)

Here, i is the current, a is area of cross-section of the wire, r is 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)

Step 3: (a) Calculations of the magnitude of the current’s magnetic field at a radial distance 0 cm

For r = 0,

Since a > 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.

Step 4: (b) Calculations of the magnitude of the current’s magnetic field at a radial distance 1.00 cm

For r=0.0100 m

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

B=μ0i2πa2r=4π×10-7T.m/A170 A2π0.0200 m20.0100 m=8.50×10-4T

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

Step 5: (c) Calculations of the magnitude of the current’s magnetic field at a radial distance 2.00 cm

For r=0.0200 m

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

B=μ0i2πa2r=4π×10-7T.m/A170 A2π0.0200 m20.0200 m=1.70×10-3T

The magnitude of the current’s magnetic field at a radial distance 2.00 cm is 1.70×10-3T.

Step 6: (d) Calculations of the magnitude of the current’s magnetic field at a radial distance 4.00 cm

For r=0.0400 m,

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

B=μ0i2πr=4π×10-7T.m/A170 A2π0.0400 m=8.50×10-4T

The magnitude of the current’s magnetic field at a radial distance 4.00 cm is 8.50×10-4T.

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