Edit to “Proof”
en/9.2.9.md
+1 −1
| ### Statement | |||
| $9.2.9.$ Prove that at large distances from two series-connected sections of wire $l_1$ and $l_2$, through which current flows, the magnetic field is close to the magnetic field of the section of wire $\vec{l} = \vec{l_1} + \vec{l_2}$ through which the same current flows. | |||
| @@ -4,6 +4,6 @@Statement | |||
| ### Proof | |||
| − | Let's find the magnetic field from each of the wires in vector form: $$ \boxed{\vec{B} = \vec{B_1} + \vec{B_2}} | ||
| + | Let's find the magnetic field from each of the wires in vector form: $$ \boxed{\vec{B} = \vec{B_1} + \vec{B_2}}\tag{1} $$ Where are the components of each magnetic induction $$ \vec{B_1} = \frac{\mu_0}{4\pi R^3}[\vec{Il_1} \times \vec{R}] = \frac{\mu_0 I}{4\pi R^3}[\vec{l_1}\times \vec{R}]\tag{2} $$ $$ \vec{B_2} = \frac{\mu_0}{4\pi R^3}[\vec{Il_2} \times \vec{R}] = \frac{\mu_0 I}{4\pi R^3}[\vec{l_2}\times \vec{R}]\tag{3} $$ We substitute the values $(2)$ and $(3)$ into $(1)$, $$ \vec{B} = \frac{\mu_0 I}{4\pi R^3}[(\vec{l_1} + \vec{l_2})\times \vec{R}] \tag{4} $$ The last transformation follows from the property of distributivity under addition for the vector product. | ||
| Thus, we have obtained that in sum the two wires have the same effect as the wire $$ \vec{l} = \vec{l_1} + \vec{l_2} $$ | |||
| ### Statement | ### Statement | ||
| $9.2.9.$ Prove that at large distances from two series-connected sections of wire $l_1$ and $l_2$, through which current flows, the magnetic field is close to the magnetic field of the section of wire $\vec{l} = \vec{l_1} + \vec{l_2}$ through which the same current flows. | $9.2.9.$ Prove that at large distances from two series-connected sections of wire $l_1$ and $l_2$, through which current flows, the magnetic field is close to the magnetic field of the section of wire $\vec{l} = \vec{l_1} + \vec{l_2}$ through which the same current flows. | ||
| @@ -4,6 +4,6 @@Statement | |||
| ### Proof | ### Proof | ||
| Let's find the magnetic field from each of the wires in vector form: $$ \boxed{\vec{B} = \vec{B_1} + \vec{B_2}} |
Let's find the magnetic field from each of the wires in vector form: $$ \boxed{\vec{B} = \vec{B_1} + \vec{B_2}}\tag{1} $$ Where are the components of each magnetic induction $$ \vec{B_1} = \frac{\mu_0}{4\pi R^3}[\vec{Il_1} \times \vec{R}] = \frac{\mu_0 I}{4\pi R^3}[\vec{l_1}\times \vec{R}]\tag{2} $$ $$ \vec{B_2} = \frac{\mu_0}{4\pi R^3}[\vec{Il_2} \times \vec{R}] = \frac{\mu_0 I}{4\pi R^3}[\vec{l_2}\times \vec{R}]\tag{3} $$ We substitute the values $(2)$ and $(3)$ into $(1)$, $$ \vec{B} = \frac{\mu_0 I}{4\pi R^3}[(\vec{l_1} + \vec{l_2})\times \vec{R}] \tag{4} $$ The last transformation follows from the property of distributivity under addition for the vector product. | ||
| Thus, we have obtained that in sum the two wires have the same effect as the wire $$ \vec{l} = \vec{l_1} + \vec{l_2} $$ | Thus, we have obtained that in sum the two wires have the same effect as the wire $$ \vec{l} = \vec{l_1} + \vec{l_2} $$ | ||