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Chapter 26, Problem 6P

Figure P26.6 represents a section of a conductor of nonuniform diameter carrying a current of I = 5.00 A. The radius of cross-section A1 is r1 = 0.400 cm. (a) What is the magnitude of the current density across A1? The radius r2 at A2 is larger than the radius r1 at A1. (b) Is the current at A2 larger, smaller, or the same? (c) Is the current density at A2 larger, smaller, or the same? Assume A2 = 4A1. Specify the (d) radius, (e) current, and (f) current density at A2.

Figure P26.6

Chapter 26, Problem 6P, Figure P26.6 represents a section of a conductor of nonuniform diameter carrying a current of I =

(a)

Expert Solution
Check Mark
To determine

The magnitude of the current density across A1 .

Answer to Problem 6P

The magnitude of the current density across A1 is 99471.83A/m2 .

Explanation of Solution

Given information: Current carried by a conductor is 5.00A , radius of cross section A1 is 0.400cm .

Write the expression for the area of cross section of a conductor.

A1=πr12 (1)

Here,

A1 is the area of cross section of a conductor.

r1 is the radius of cross section A1 .

Substitute 0.400cm for r1 in equation (1) to find A1 ,

A1=π(0.400cm×1m100cm)2=5.026×105m2

Thus, the area of cross section of a conductor is 5.026×105m2 .

Formula to calculate the current density across A1 .

J1=I1A1 (2)

Here,

J1 is the current density across A1 .

I1 is the current carried by a conductor.

Substitute 5.00A for I1 , 5.026×105m2 for A1 in equation (2) to find J1 ,

J1=5.00A5.026×105m2=99471.83A/m2

Thus, the magnitude of the current density across A1 is 99471.83A/m2 .

Conclusion:

Therefore, the magnitude of the current density across A1 is 99471.83A/m2 .

(b)

Expert Solution
Check Mark
To determine

The reason that the current at A2 is smaller, larger or remains same.

Answer to Problem 6P

The current at A2 remains same because the current is not depending on cross sectional area. So, the current at A2 is equal to the current at A1 .

Explanation of Solution

Given information: Current carried by a conductor is 5.00A , radius of cross section A1 is 0.400cm .

The current is not depending on cross sectional area. So, the current at A2 is equal to the current at A1 . Therefore, the current is the same.

Thus, the density at A2 remains same because the current is not depending on cross sectional area. So, the current at A2 is equal to the current at A1 .

Conclusion:

Therefore, the density at A2 remains same because the current is not depending on cross sectional

area. So, the current at A2 is equal to the current at A1 .

(c)

Expert Solution
Check Mark
To determine

The reason that the current density at A2 is smaller, larger or remains same.

Answer to Problem 6P

The current density at A2 is smaller than the current density at A1 because area of cross section at A2 is greater than the area of cross section at A1 .

Explanation of Solution

Given information: Current carried by a conductor is 5.00A , radius of cross section A1 is 0.400cm .

From equation (2),

Formula to calculate the current density across A1 .

J1=I1A1

Here,

J1 is the current density across A1 .

Formula to calculate the current density across A2 .

J2=I2A2

Here,

J2 is the current density across A2 .

I2 is the current carried by a conductor.

From above relations, the current density is inversely proportional to area of cross section. From the figure given in the question, it is shown that:

A2>A1

Hence, the current density at A2 is smaller than the current density at A1 .

J2>J1

Thus, the current density at A2 is smaller than the current density at A1 because area of cross section at A2 is greater than the area of cross section at A1 .

Conclusion:

Therefore, the current density at A2 is smaller than the current density at A1 because area of cross section at A2 is greater than the area of cross section at A1 .

(d)

Expert Solution
Check Mark
To determine

The radius of cross section at A2 .

Answer to Problem 6P

The radius of cross section at A2 is 8.00mm .

Explanation of Solution

Given information: Current carried by a conductor is 5.00A , radius of cross section A1 is 0.400cm , assume A2=4A1 .

It is given that the expression for the crossectional area is:

A2=4A1 (3)

From equation (1),

Write the expression for the area of cross section of a conductor A1 .

A1=πr12

Write the expression for the area of cross section of a conductor A2 .

A2=πr22

Here,

A2 is the area of cross section of a conductor.

r2 is the radius of cross section A2 .

Substitute πr12 for r1 , πr22 for A2 in equation (3) to find r2 ,

πr22=4(πr12)r22=4r12r2=4r12=2r1 (4)

Substitute 0.400cm for r1 in equation (4) to find r2 ,

Thus, the area of cross section of a conductor is 5.026×105m2 .

r2=2×(0.400cm×1m100cm)=8×103m=8.00mm

Thus, the radius of cross section at A2 is 8.00mm .

Conclusion:

Therefore, the radius of cross section at A2 is 8.00mm .

(e)

Expert Solution
Check Mark
To determine

The current for cross section at A2 .

Answer to Problem 6P

The current for cross section at A2 is 5.00A .

Explanation of Solution

Given information: Current carried by a conductor is 5.00A , radius of cross section A1 is 0.400cm .

The current is not depending on cross sectional area. So, the current at A2 is equal to the current at A1 . Therefore, the current is the same as for cross section 1.

I2=I1=5.00A

Thus, the current for cross section at A2 is 5.00A .

Conclusion:

Therefore, the current for cross section at A2 is 5.00A .

(f)

Expert Solution
Check Mark
To determine

The magnitude of the current density across A2 .

Answer to Problem 6P

The magnitude of the current density across A2 is 24875.62A/m2 .

Explanation of Solution

Given information: Current carried by a conductor is 5.00A , radius of cross section A1 is 0.400cm .

Write the expression for the area of cross section of a conductor A2 .

A2=πr22 (5)

Here,

A2 is the area of cross section of a conductor.

r2 is the radius of cross section A2 .

Substitute 8.00mm for r1 in equation (5) to find A2 ,

A2=π(8.00mm×1m1000mm)2=2.010×104m2

Thus, the area of cross section of a conductor is 2.010×104m2 .

Formula to calculate the current density across A2 .

J2=I2A2 (6)

Substitute 5.00A for I2 , 2.010×104m2 for A2 in equation (6) to find J2 ,

J2=5.00A2.010×104m2=24875.62A/m2

Thus, the magnitude of the current density across A2 is 24875.62A/m2 .

Conclusion:

Therefore, the magnitude of the current density across A2 is 24875.62A/m2 .

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Chapter 26 Solutions

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term

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