Правка раздела «Solution»
en/9.2.6.md
+1 −1
| ### Statement | |||
| $9.2.6.$ Long straight wires with current intersect at right angles. Determine the magnetic field induction at a point with coordinates $x$ and $y$ if the coordinate axes are the wires and the current in the wires is $I$. | |||
| ### Solution | |||
| @@ -6,8 +6,8 @@Solution | |||
|  | |||
| − | We find the total vector of magnetic induction as the sum of the vectors of magnetic induction created by each of the vectors $$ \vec{B} = \vec{B_1} + \vec{B_2} $$ Using the right-hand rule, we can determine the direction of the magnetic induction lines and notice that $\vec{B_1}$ and $\vec{B_2}$ are directed perpendicular to the $xy$ plane, which means they are co-directed $$ \boxed{B=B_1+B_2} | ||
| + | We find the total vector of magnetic induction as the sum of the vectors of magnetic induction created by each of the vectors $$ \vec{B} = \vec{B_1} + \vec{B_2} $$ Using the right-hand rule, we can determine the direction of the magnetic induction lines and notice that $\vec{B_1}$ and $\vec{B_2}$ are directed perpendicular to the $xy$ plane, which means they are co-directed $$ \boxed{B=B_1+B_2}\tag{1} $$ We will find the magnetic induction of each of them as the induction of an infinite single-horned wire. $$ B_1 = \frac{\mu_0 I}{2\pi} \cdot \frac{1}{x};\quad B_2 = \frac{\mu_0 I}{2\pi} \cdot \frac{1}{y}\tag{2} $$ We substitute the obtained expression $(2)$ into $(1)$ $$ \boxed{B = {B_1} +{B_2} = \frac{\mu_0 I}{2\pi}\cdot \left(\frac{1}{x}+\frac{1}{y}\right)} $$ | ||
| #### Answer | |||
| $$B=\frac{\mu_0 I}{2\pi}\cdot \left(\frac{1}{x}+\frac{1}{y}\right)$$ | |||
| ### Statement | ### Statement | ||
| $9.2.6.$ Long straight wires with current intersect at right angles. Determine the magnetic field induction at a point with coordinates $x$ and $y$ if the coordinate axes are the wires and the current in the wires is $I$. | $9.2.6.$ Long straight wires with current intersect at right angles. Determine the magnetic field induction at a point with coordinates $x$ and $y$ if the coordinate axes are the wires and the current in the wires is $I$. | ||
| ### Solution | ### Solution | ||
| @@ -6,8 +6,8 @@Solution | |||
|  |  | ||
| We find the total vector of magnetic induction as the sum of the vectors of magnetic induction created by each of the vectors $$ \vec{B} = \vec{B_1} + \vec{B_2} $$ Using the right-hand rule, we can determine the direction of the magnetic induction lines and notice that $\vec{B_1}$ and $\vec{B_2}$ are directed perpendicular to the $xy$ plane, which means they are co-directed $$ \boxed{B=B_1+B_2} |
We find the total vector of magnetic induction as the sum of the vectors of magnetic induction created by each of the vectors $$ \vec{B} = \vec{B_1} + \vec{B_2} $$ Using the right-hand rule, we can determine the direction of the magnetic induction lines and notice that $\vec{B_1}$ and $\vec{B_2}$ are directed perpendicular to the $xy$ plane, which means they are co-directed $$ \boxed{B=B_1+B_2}\tag{1} $$ We will find the magnetic induction of each of them as the induction of an infinite single-horned wire. $$ B_1 = \frac{\mu_0 I}{2\pi} \cdot \frac{1}{x};\quad B_2 = \frac{\mu_0 I}{2\pi} \cdot \frac{1}{y}\tag{2} $$ We substitute the obtained expression $(2)$ into $(1)$ $$ \boxed{B = {B_1} +{B_2} = \frac{\mu_0 I}{2\pi}\cdot \left(\frac{1}{x}+\frac{1}{y}\right)} $$ | ||
| #### Answer | #### Answer | ||
| $$B=\frac{\mu_0 I}{2\pi}\cdot \left(\frac{1}{x}+\frac{1}{y}\right)$$ | $$B=\frac{\mu_0 I}{2\pi}\cdot \left(\frac{1}{x}+\frac{1}{y}\right)$$ | ||