Chapter 20, Problem 44P

(a)

To determine

The magnetic field magnitude due to wire A at the position of wire B.

Answer to Problem 44P

The magnetic field due to wire A at the position of wire B is $2.7\times {10}^{-6}\text{T}$

Explanation of Solution

Given:

Current in the wire A is 2 A and in wire B is 4 A flowing in the same direction. Both the wires are 15 cm apart.

Formula used:

The magnetic field is calculated as

  $B=\frac{{\mu }_0}{2\pi }\frac{I}{r}$

Calculation:

The magnetic field due to wire A at the position of wire B is

  $\begin{array}{l}{B}_{A to B}=\frac{{\mu }_0}{2\pi }\frac{I_A}{r_{A to B}}\\ B_{A to B}=\frac{(4\pi \times {10}^{-7})(2)}{2\pi (0.15)}=2.667\times {10}^{-6}\text{T}\\ B_{A to B}\approx 2.7\times {10}^{-6}\text{T}\end{array}$

Conclusion:

The magnetic field due to wire A at the position of wire B is $2.7\times {10}^{-6}\text{T}$

(b)

To determine

The magnetic field due to wire B at the position of wire A.

Answer to Problem 44P

The magnetic field due to wire B at the position of wire A is $5.3\times {10}^{-6}\text{T}$

Explanation of Solution

Given:

Current in the wire A is 2 A and in wire B is 4 A flowing in the same direction. Both the wires are 15 cm apart.

Formula used:

The magnetic field is calculated as

  $B=\frac{{\mu }_0}{2\pi }\frac{I}{r}$

Calculation:

The magnetic field due to wire B at the position of wire A is

  $\begin{array}{l}{B}_{B to A}=\frac{{\mu }_0}{2\pi }\frac{I_B}{r_{B to A}}\\ B_{B to A}=\frac{{\text{(4π×10}}^{\text{-7}}\text{T}\text{.m/A)(4A)}}{\text{2π(0}\text{.15m)}}=5.333\times {10}^{-6}\text{T}\\ B_{B to A}\approx 5.3\times {10}^{-6}\text{T}\end{array}$

Conclusion:

The magnetic field due to wire A at the position of wire B is $5.3\times {10}^{-6}\text{T}$

(c)

To determine

To identify: Whether the two magnetic fields will equal and opposite.

Answer to Problem 44P

The two magnetic fields are not equal and opposite.

Explanation of Solution

Introduction:

When a wire carries current then the movement of charges induces magnetic field in the wire.

Here, both wires carry a separate value of the current and magnetic field. Every wire that carries movement of charges i.e. current, it induces its separate magnetic field. It does not depend on the magnetic field caused by another wire.

Conclusion:

Thus, the magnetic field induces by separate wires does not have equal magnitudes and certainly they are not opposite because they have not direct relation whatsoever.

(d)

To determine

The force on wire A due to wire B and the force on wire B due to wire A.

To identify: Whether these are forces equal and opposite.

Answer to Problem 44P

The force on wire A due to wire B and force on wire B due to wire A is same and equal to $1.1\times {10}^{-5}\text{N/m}$ . The two magnetic forces are equal and opposite.

Explanation of Solution

Given:

Current in the wire A is 2 A and in wire B is 4 A flowing in the same direction. Both the wires are 15 cm apart.

Formula used:

The magnetic force is calculated as

  $F=\frac{{\mu }_0}{2\pi }\frac{I_A{I}_B}{d_{A to B}}{l}_B$

Calculation:

Since both the wires carry current in the same direction. The magnetic force on wire B due to wire A and on wire B due to wire A will have equal magnitude.

  $\begin{array}{c}\frac{F_{onAduetoB}}{l_A}=\frac{F_{on\text{ B }dueto\text{ A}}}{l_B}\\ =\frac{{\mu }_0}{2\pi }\frac{I_A{I}_B}{d_{A to B}}\\ \text{=}\frac{{\text{(4π×10}}^{\text{-7}}\text{T}\text{.m/A)(2A)(4A)}}{\text{2π(0}\text{.15m)}}\\ \text{=1}{\text{.067×10}}^{\text{-5}}\text{N/m}\\ \approx \text{1}{\text{.1×10}}^{\text{-5}}\text{N/m}\end{array}$

These two forces per unit length will have equal magnitude but they will act in the opposite direction because they are a pair of Newton’s third law.

Conclusion:

The force on wire A due to wire B and on wire B due to wire A is same and equal to $1.1\times {10}^{-5}\text{N/m}$ . The two magnetic forces are equal and opposite.