Chapter 20, Problem 37P

(a)

To determine

The magnetic field midway between two long straight current-carrying wires for the given condition.

Answer to Problem 37P

The net magnetic field in between the current-carrying wires flowing in the same direction is

  $(2\times {10}^{-5})(I-15)\text{T}$ .

Explanation of Solution

Given:

Two long straight wires are 2 cm apart. Current in one wire is I and other carries a current of 15 A in same direction.

Formula used:

Magnetic field in the current-carrying wire is calculated as

  $B=\frac{{\mu }_{0}I}{2\pi r}$

Calculation:

If the current in the two wires is in the same direction then the magnetic field at the midway to both the wires will oppose each other and so the net magnetic field will be the difference of both.

  $\begin{array}{l}{B}_{net}=\frac{{\mu }_0{I}_1}{2\pi r_1}-\frac{{\mu }_0{I}_2}{2\pi r_2}\\ B_{net}=\frac{(4\pi \times {10}^{-7})}{2\pi (1\times {10}^{-2})}(I-15)\\ B_{net}=(2\times {10}^{-5})(I-15)\end{array}$

Conclusion:

The net magnetic field in between the current-carrying wires flowing in the same direction is $(2\times {10}^{-5})(I-15)\text{T}$ .

(b)

To determine

The magnetic field midway between two long straight current-carrying wires for the given condition.

Answer to Problem 37P

The net magnetic field in between the current-carrying wires flowing in the opposite direction is

  $(2\times {10}^{-5})(I+15)\text{T}$ .

Explanation of Solution

Given:

Two long straight wires are 2 cm apart. Current in one wire is I and other carries a current of 15 A in opposite direction.

Formula used:

Magnetic field in the current-carrying wire is calculated as

  $B=\frac{{\mu }_{0}I}{2\pi r}$

Calculation:

If the current in the two wires is in the opposite direction then the magnetic field at the midway to both the wires will attract each other and so the net magnetic field will be the sum of both.

  $\begin{array}{l}{B}_{net}=\frac{{\mu }_0{I}_1}{2\pi r_1}+\frac{{\mu }_0{I}_2}{2\pi r_2}\\ B_{net}=\frac{(4\pi \times {10}^{-7})}{2\pi (1\times {10}^{-2})}(I+15)\\ B_{net}=(2\times {10}^{-5})(I+15)\text{T}\end{array}$

Conclusion:

The net magnetic field in between the current-carrying wires flowing in the opposite direction is $(2\times {10}^{-5})(I+15)\text{T}$ .