Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 22, Problem 45P

(a)

To determine

The magnitude of the magnetic field created by I1 at the location of  I2 .

(a)

Expert Solution
Check Mark

Answer to Problem 45P

The magnitude of the magnetic field created by I1 at the location I2 is 10μT .

Explanation of Solution

Write the expression of the magnetic field at the origin due to the current I1 .

  B1=μ0I12πa=(μ04π)(2I1a)        (I)

Here, a is the distance between the wire and the origin, μ0 is the permeability of free space, and I1 current in the first wire.

Conclusion:

Substitute 1×107N/A2  for μ0/4π, 5.00A for I110.0cm for a in the above expression (I),

    B1=(1×107N/A210.0cm)(2×5.00A1m100cm)=1×105T(1μT106T)=10μT                                       

The magnitude of the magnetic field created by I1 at the location I2 is 10μT .

(b)

To determine

The force per unit length

(b)

Expert Solution
Check Mark

Answer to Problem 45P

The force per unit length exerted by I1 on I2 is 80μN towards the first wire.

Explanation of Solution

Write the expression of the force exerted per unit length by I1 on I2 .

  F1=B1I2        (I)

Here, F1 is the force per unit length, B1 is the magnetic field due to the current I1 on I2 .

Conclusion:

Substitute 1×105T  for B1, 8.00A for I2,  in the above expression (I),

    F1=(1×105T)(8.00A)=8×105T×106μN1N=80μN                                       

The force per unit length exerted by I1 on I2 is 80μN towards the first wire.

(c)

To determine

The magnitude of the magnetic field created by I2 at the location of  I1 .

(c)

Expert Solution
Check Mark

Answer to Problem 45P

The magnitude of the magnetic field created by I2 at the location I1 is 16μT .

Explanation of Solution

Write the expression of the magnetic field at the origin due to the current I1 .

  B1=μ0I12πa=(μ04π)(2I1a)        (I)

Here, a is the distance between the wire and the origin, μ0 is the permeability of free space, and I1 current in the first wire.

Conclusion:

Substitute 1×107N/A2  for μ0/4π, 8.00A for I210.0cm for a in the above expression (I),

    B1=(1×107N/A210.0cm)(2×8.00A1m100cm)=1×105T(106μT1T)=16μT                                       

The magnitude of the magnetic field created by I2 at the location I1 is 16μT .

(d)

To determine

The force per unit length

(d)

Expert Solution
Check Mark

Answer to Problem 45P

The force per unit length exerted by I2 on I1 is 80μN towards the first wire.

Explanation of Solution

Write the expression of the force exerted per unit length by I2 on I1 .

  F2=B1I1        (I)

Here, F2 is the force per unit length, B1 is the magnetic field due to the current I1 on I2 .

Conclusion:

Substitute 1×105T  for B1, 5.00A for I1,  in the above expression (I),

    F1=(1.6×105T)(5.00A)=8×105T×106μN1N=80μN                                       

The force per unit length exerted by I2 on I1 is 80μN towards the first wire.

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Chapter 22 Solutions

Principles of Physics: A Calculus-Based Text

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