The net current through the loop that must be present.

Question

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Answer 1

Question

Chapter 31, Problem 58PQ

To determine

The net current through the loop that must be present.

Expert Solution & Answer

Answer to Problem 58PQ

The net current through the loop that must be present is 124.34 A_.

Explanation of Solution

Write the expression for the Ampere circuital law as.

∮B⋅dl=μ0I

Rearrange the above equation for I.

I=∮B.dlμ0 (I)

Here, B is the magnetic field , l is length of the wire, μ0 is the permeability of free space and I is the current flowing through the wire.

The contribution to the circulation integral due to each segment of loop is.

Circulation integral=∮B⋅dl (II)

The two segments of the square loop are parallel to the magnetic field. Therefore, the contribution to circulation integral due to these two segments is zero.

Write the expression for the total contribution to the circulation integral as.

∮B⋅dl=(∮B⋅dl)x+(∮B⋅dl)y (III)

Here, (∮B⋅dl)x is the circulation integral in the x-direction, (∮B⋅dl)y is the circulation integral in the y-direction and ∮B⋅dl is the total circulation integral.

Conclusion:

The length of the segment in the X-direction is 2.50 cm.

Substitute (∮B⋅dl)x for ∮B⋅dl , (0.125y i^−0.125x j^)T for B and dx i^ for dl in equation (II).

(∮B⋅dl)x=∮{(0.125y i^−0.125x j^)T⋅dx i^}=0.125y∫02.50dx=0.125y[x]02.50=0.125y(2.50 cm)

Substitute 2.50 cm for y in above equation.

(∮B⋅dl)x=0.125(2.50 cm(1 m100 cm))(2.50 cm(1 m100 cm))=7.8125×10−5 T⋅m

The length of the segment in the Y-direction is 2.50 cm.

Substitute (∮B⋅dl)y for ∮B⋅dl , (0.125y i^−0.125x j^)T for B and dy j^ for dl in equation (II).

(∮B⋅dl)y=∮{(0.125y i^−0.125x j^)T⋅dy j^}=0.125x∫02.50dy=0.125x[y]02.50=0.125x(2.50 cm)

Substitute 2.50 cm for x in above equation.

(∮B⋅dl)y=0.125(2.50 cm(1 m100 cm))(2.50 cm(1 m100 cm))=7.8125×10−5 T⋅m

Substitute 7.8125×10−5 T⋅m for (∮B⋅dl)x and 7.8125×10−5 T⋅m(∮B⋅dl)y in equation (III).

∮B⋅dl=(7.8125×10−5 T⋅m)+(7.8125×10−5 T⋅m)=1.5625×10−4 T⋅m

Substitute 1.5625×10−4 T⋅m for ∮B⋅dl and 4π×10−7 T⋅mA for μ0 in equation (I).

I=(1.5625×10−4 T⋅m)(4π×10−7 T⋅mA)=124.34 A

Thus, the net current through the loop that must be present is 124.34 A_.

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Answer 2

Question

Chapter 31, Problem 58PQ

To determine

The net current through the loop that must be present.

Expert Solution & Answer

Answer to Problem 58PQ

The net current through the loop that must be present is 124.34 A_.

Explanation of Solution

Write the expression for the Ampere circuital law as.

∮B⋅dl=μ0I

Rearrange the above equation for I.

I=∮B.dlμ0 (I)

Here, B is the magnetic field , l is length of the wire, μ0 is the permeability of free space and I is the current flowing through the wire.

The contribution to the circulation integral due to each segment of loop is.

Circulation integral=∮B⋅dl (II)

The two segments of the square loop are parallel to the magnetic field. Therefore, the contribution to circulation integral due to these two segments is zero.

Write the expression for the total contribution to the circulation integral as.

∮B⋅dl=(∮B⋅dl)x+(∮B⋅dl)y (III)

Here, (∮B⋅dl)x is the circulation integral in the x-direction, (∮B⋅dl)y is the circulation integral in the y-direction and ∮B⋅dl is the total circulation integral.

Conclusion:

The length of the segment in the X-direction is 2.50 cm.

Substitute (∮B⋅dl)x for ∮B⋅dl , (0.125y i^−0.125x j^)T for B and dx i^ for dl in equation (II).

(∮B⋅dl)x=∮{(0.125y i^−0.125x j^)T⋅dx i^}=0.125y∫02.50dx=0.125y[x]02.50=0.125y(2.50 cm)

Substitute 2.50 cm for y in above equation.

(∮B⋅dl)x=0.125(2.50 cm(1 m100 cm))(2.50 cm(1 m100 cm))=7.8125×10−5 T⋅m

The length of the segment in the Y-direction is 2.50 cm.

Substitute (∮B⋅dl)y for ∮B⋅dl , (0.125y i^−0.125x j^)T for B and dy j^ for dl in equation (II).

(∮B⋅dl)y=∮{(0.125y i^−0.125x j^)T⋅dy j^}=0.125x∫02.50dy=0.125x[y]02.50=0.125x(2.50 cm)

Substitute 2.50 cm for x in above equation.

(∮B⋅dl)y=0.125(2.50 cm(1 m100 cm))(2.50 cm(1 m100 cm))=7.8125×10−5 T⋅m

Substitute 7.8125×10−5 T⋅m for (∮B⋅dl)x and 7.8125×10−5 T⋅m(∮B⋅dl)y in equation (III).

∮B⋅dl=(7.8125×10−5 T⋅m)+(7.8125×10−5 T⋅m)=1.5625×10−4 T⋅m

Substitute 1.5625×10−4 T⋅m for ∮B⋅dl and 4π×10−7 T⋅mA for μ0 in equation (I).

I=(1.5625×10−4 T⋅m)(4π×10−7 T⋅mA)=124.34 A

Thus, the net current through the loop that must be present is 124.34 A_.

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Answer 3

Question

Chapter 31, Problem 58PQ

To determine

The net current through the loop that must be present.

Expert Solution & Answer

Answer to Problem 58PQ

The net current through the loop that must be present is 124.34 A_.

Explanation of Solution

Write the expression for the Ampere circuital law as.

∮B⋅dl=μ0I

Rearrange the above equation for I.

I=∮B.dlμ0 (I)

Here, B is the magnetic field , l is length of the wire, μ0 is the permeability of free space and I is the current flowing through the wire.

The contribution to the circulation integral due to each segment of loop is.

Circulation integral=∮B⋅dl (II)

The two segments of the square loop are parallel to the magnetic field. Therefore, the contribution to circulation integral due to these two segments is zero.

Write the expression for the total contribution to the circulation integral as.

∮B⋅dl=(∮B⋅dl)x+(∮B⋅dl)y (III)

Here, (∮B⋅dl)x is the circulation integral in the x-direction, (∮B⋅dl)y is the circulation integral in the y-direction and ∮B⋅dl is the total circulation integral.

Conclusion:

The length of the segment in the X-direction is 2.50 cm.

Substitute (∮B⋅dl)x for ∮B⋅dl , (0.125y i^−0.125x j^)T for B and dx i^ for dl in equation (II).

(∮B⋅dl)x=∮{(0.125y i^−0.125x j^)T⋅dx i^}=0.125y∫02.50dx=0.125y[x]02.50=0.125y(2.50 cm)

Substitute 2.50 cm for y in above equation.

(∮B⋅dl)x=0.125(2.50 cm(1 m100 cm))(2.50 cm(1 m100 cm))=7.8125×10−5 T⋅m

The length of the segment in the Y-direction is 2.50 cm.

Substitute (∮B⋅dl)y for ∮B⋅dl , (0.125y i^−0.125x j^)T for B and dy j^ for dl in equation (II).

(∮B⋅dl)y=∮{(0.125y i^−0.125x j^)T⋅dy j^}=0.125x∫02.50dy=0.125x[y]02.50=0.125x(2.50 cm)

Substitute 2.50 cm for x in above equation.

(∮B⋅dl)y=0.125(2.50 cm(1 m100 cm))(2.50 cm(1 m100 cm))=7.8125×10−5 T⋅m

Substitute 7.8125×10−5 T⋅m for (∮B⋅dl)x and 7.8125×10−5 T⋅m(∮B⋅dl)y in equation (III).

∮B⋅dl=(7.8125×10−5 T⋅m)+(7.8125×10−5 T⋅m)=1.5625×10−4 T⋅m

Substitute 1.5625×10−4 T⋅m for ∮B⋅dl and 4π×10−7 T⋅mA for μ0 in equation (I).

I=(1.5625×10−4 T⋅m)(4π×10−7 T⋅mA)=124.34 A

Thus, the net current through the loop that must be present is 124.34 A_.

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Answer 4

Question

Chapter 31, Problem 58PQ

To determine

The net current through the loop that must be present.

Expert Solution & Answer

Answer to Problem 58PQ

The net current through the loop that must be present is 124.34 A_.

Explanation of Solution

Write the expression for the Ampere circuital law as.

∮B⋅dl=μ0I

Rearrange the above equation for I.

I=∮B.dlμ0 (I)

Here, B is the magnetic field , l is length of the wire, μ0 is the permeability of free space and I is the current flowing through the wire.

The contribution to the circulation integral due to each segment of loop is.

Circulation integral=∮B⋅dl (II)

The two segments of the square loop are parallel to the magnetic field. Therefore, the contribution to circulation integral due to these two segments is zero.

Write the expression for the total contribution to the circulation integral as.

∮B⋅dl=(∮B⋅dl)x+(∮B⋅dl)y (III)

Here, (∮B⋅dl)x is the circulation integral in the x-direction, (∮B⋅dl)y is the circulation integral in the y-direction and ∮B⋅dl is the total circulation integral.

Conclusion:

The length of the segment in the X-direction is 2.50 cm.

Substitute (∮B⋅dl)x for ∮B⋅dl , (0.125y i^−0.125x j^)T for B and dx i^ for dl in equation (II).

(∮B⋅dl)x=∮{(0.125y i^−0.125x j^)T⋅dx i^}=0.125y∫02.50dx=0.125y[x]02.50=0.125y(2.50 cm)

Substitute 2.50 cm for y in above equation.

(∮B⋅dl)x=0.125(2.50 cm(1 m100 cm))(2.50 cm(1 m100 cm))=7.8125×10−5 T⋅m

The length of the segment in the Y-direction is 2.50 cm.

Substitute (∮B⋅dl)y for ∮B⋅dl , (0.125y i^−0.125x j^)T for B and dy j^ for dl in equation (II).

(∮B⋅dl)y=∮{(0.125y i^−0.125x j^)T⋅dy j^}=0.125x∫02.50dy=0.125x[y]02.50=0.125x(2.50 cm)

Substitute 2.50 cm for x in above equation.

(∮B⋅dl)y=0.125(2.50 cm(1 m100 cm))(2.50 cm(1 m100 cm))=7.8125×10−5 T⋅m

Substitute 7.8125×10−5 T⋅m for (∮B⋅dl)x and 7.8125×10−5 T⋅m(∮B⋅dl)y in equation (III).

∮B⋅dl=(7.8125×10−5 T⋅m)+(7.8125×10−5 T⋅m)=1.5625×10−4 T⋅m

Substitute 1.5625×10−4 T⋅m for ∮B⋅dl and 4π×10−7 T⋅mA for μ0 in equation (I).

I=(1.5625×10−4 T⋅m)(4π×10−7 T⋅mA)=124.34 A

Thus, the net current through the loop that must be present is 124.34 A_.

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Answer 5

Chapter 31, Problem 58PQ

To determine

The net current through the loop that must be present.

Expert Solution & Answer

Answer to Problem 58PQ

The net current through the loop that must be present is 124.34 A_.

Explanation of Solution

Write the expression for the Ampere circuital law as.

∮B⋅dl=μ0I

Rearrange the above equation for I.

I=∮B.dlμ0 (I)

Here, B is the magnetic field , l is length of the wire, μ0 is the permeability of free space and I is the current flowing through the wire.

The contribution to the circulation integral due to each segment of loop is.

Circulation integral=∮B⋅dl (II)

The two segments of the square loop are parallel to the magnetic field. Therefore, the contribution to circulation integral due to these two segments is zero.

Write the expression for the total contribution to the circulation integral as.

∮B⋅dl=(∮B⋅dl)x+(∮B⋅dl)y (III)

Here, (∮B⋅dl)x is the circulation integral in the x-direction, (∮B⋅dl)y is the circulation integral in the y-direction and ∮B⋅dl is the total circulation integral.

Conclusion:

The length of the segment in the X-direction is 2.50 cm.

Substitute (∮B⋅dl)x for ∮B⋅dl , (0.125y i^−0.125x j^)T for B and dx i^ for dl in equation (II).

(∮B⋅dl)x=∮{(0.125y i^−0.125x j^)T⋅dx i^}=0.125y∫02.50dx=0.125y[x]02.50=0.125y(2.50 cm)

Substitute 2.50 cm for y in above equation.

(∮B⋅dl)x=0.125(2.50 cm(1 m100 cm))(2.50 cm(1 m100 cm))=7.8125×10−5 T⋅m

The length of the segment in the Y-direction is 2.50 cm.

Substitute (∮B⋅dl)y for ∮B⋅dl , (0.125y i^−0.125x j^)T for B and dy j^ for dl in equation (II).

(∮B⋅dl)y=∮{(0.125y i^−0.125x j^)T⋅dy j^}=0.125x∫02.50dy=0.125x[y]02.50=0.125x(2.50 cm)

Substitute 2.50 cm for x in above equation.

(∮B⋅dl)y=0.125(2.50 cm(1 m100 cm))(2.50 cm(1 m100 cm))=7.8125×10−5 T⋅m

Substitute 7.8125×10−5 T⋅m for (∮B⋅dl)x and 7.8125×10−5 T⋅m(∮B⋅dl)y in equation (III).

∮B⋅dl=(7.8125×10−5 T⋅m)+(7.8125×10−5 T⋅m)=1.5625×10−4 T⋅m

Substitute 1.5625×10−4 T⋅m for ∮B⋅dl and 4π×10−7 T⋅mA for μ0 in equation (I).

I=(1.5625×10−4 T⋅m)(4π×10−7 T⋅mA)=124.34 A

Thus, the net current through the loop that must be present is 124.34 A_.

Answer 6

Chapter 31, Problem 58PQ

To determine

The net current through the loop that must be present.

Expert Solution & Answer

Answer to Problem 58PQ

The net current through the loop that must be present is 124.34 A_.

Explanation of Solution

Write the expression for the Ampere circuital law as.

∮B⋅dl=μ0I

Rearrange the above equation for I.

I=∮B.dlμ0 (I)

Here, B is the magnetic field , l is length of the wire, μ0 is the permeability of free space and I is the current flowing through the wire.

The contribution to the circulation integral due to each segment of loop is.

Circulation integral=∮B⋅dl (II)

The two segments of the square loop are parallel to the magnetic field. Therefore, the contribution to circulation integral due to these two segments is zero.

Write the expression for the total contribution to the circulation integral as.

∮B⋅dl=(∮B⋅dl)x+(∮B⋅dl)y (III)

Here, (∮B⋅dl)x is the circulation integral in the x-direction, (∮B⋅dl)y is the circulation integral in the y-direction and ∮B⋅dl is the total circulation integral.

Conclusion:

The length of the segment in the X-direction is 2.50 cm.

Substitute (∮B⋅dl)x for ∮B⋅dl , (0.125y i^−0.125x j^)T for B and dx i^ for dl in equation (II).

(∮B⋅dl)x=∮{(0.125y i^−0.125x j^)T⋅dx i^}=0.125y∫02.50dx=0.125y[x]02.50=0.125y(2.50 cm)

Substitute 2.50 cm for y in above equation.

(∮B⋅dl)x=0.125(2.50 cm(1 m100 cm))(2.50 cm(1 m100 cm))=7.8125×10−5 T⋅m

The length of the segment in the Y-direction is 2.50 cm.

Substitute (∮B⋅dl)y for ∮B⋅dl , (0.125y i^−0.125x j^)T for B and dy j^ for dl in equation (II).

(∮B⋅dl)y=∮{(0.125y i^−0.125x j^)T⋅dy j^}=0.125x∫02.50dy=0.125x[y]02.50=0.125x(2.50 cm)

Substitute 2.50 cm for x in above equation.

(∮B⋅dl)y=0.125(2.50 cm(1 m100 cm))(2.50 cm(1 m100 cm))=7.8125×10−5 T⋅m

Substitute 7.8125×10−5 T⋅m for (∮B⋅dl)x and 7.8125×10−5 T⋅m(∮B⋅dl)y in equation (III).

∮B⋅dl=(7.8125×10−5 T⋅m)+(7.8125×10−5 T⋅m)=1.5625×10−4 T⋅m

Substitute 1.5625×10−4 T⋅m for ∮B⋅dl and 4π×10−7 T⋅mA for μ0 in equation (I).

I=(1.5625×10−4 T⋅m)(4π×10−7 T⋅mA)=124.34 A

Thus, the net current through the loop that must be present is 124.34 A_.

Answer 7

Chapter 31, Problem 58PQ

To determine

The net current through the loop that must be present.

Answer 8

Expert Solution & Answer

Answer to Problem 58PQ

The net current through the loop that must be present is 124.34 A_.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write the expression for the Ampere circuital law as.

∮B⋅dl=μ0I

Rearrange the above equation for I.

I=∮B.dlμ0 (I)

Here, B is the magnetic field , l is length of the wire, μ0 is the permeability of free space and I is the current flowing through the wire.

The contribution to the circulation integral due to each segment of loop is.

Circulation integral=∮B⋅dl (II)

The two segments of the square loop are parallel to the magnetic field. Therefore, the contribution to circulation integral due to these two segments is zero.

Write the expression for the total contribution to the circulation integral as.

∮B⋅dl=(∮B⋅dl)x+(∮B⋅dl)y (III)

Here, (∮B⋅dl)x is the circulation integral in the x-direction, (∮B⋅dl)y is the circulation integral in the y-direction and ∮B⋅dl is the total circulation integral.