Edit to “Statement”
en/9.1.3.md
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| ### Statement | |||
| $9.1.3.$ A linear conductor of length $l$, located perpendicular to a magnetic field, is acted upon by a force $F$ if the current in the conductor is $I$. With what force will the magnetic field act on: | |||
| @@ -5,7 +5,7 @@Statement | |||
| a) a conductor of length $l + L$ bent at an angle $\varphi$, if the plane of the bend is perpendicular to the magnetic field, and the current in the conductor is $I_1$ | |||
| b) a conductor in the form of a semicircle of radius $R$, through which a current $I_2$ flows, if the plane of the semicircle is perpendicular to the magnetic field? | |||
| − |  | ||
| ### Solution | |||
| a) From Ampere's law, we find the magnetic field induction $B$ $$ F_1 = Il \cdot B \Rightarrow B = \frac{F}{Il}\tag{1} $$ in this case, the numbers acting on the part of the wire with currents $I_1$ and $I_2$, respectively $$ F_1 = I_1l \cdot B; \quad F_2 = I_1L\cdot B $$ We substitute $(1)$: $$ F_1= \frac{I_1}{I}F; \quad F_2 = \frac{I_1L}{Il}F $$ The force acting on a piece of wire bent at an angle $\varphi$ can be found from the cosine law $$ |\vec{F_p}| = |\vec{F_1} + \vec{F_2}| = \sqrt{{F_1}^2+{F_2}^2+2 F_1 F_2\cdot\cos\varphi} $$ $$ \boxed{|\vec{F_p}| = F\frac{I_1}{I}\sqrt{1+\frac{L^2}{l^2} + 2\frac{L}{l}\cdot\cos\varphi}} $$ b) For a conductor made of a semicircle, we will use the same formula for the magnetic field induction $B$ $$ B = \frac{F}{Il} $$ On a small part of a conductor of length $dl$, the Ampere force acts $$ dF = B I_2~dl = B I_2R~d\theta $$ To find the force $F$, we integrate the forces acting on each part of the semiring. $$ F = \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\theta dF = BI_2R\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\theta~d\theta $$ After mathematical transformations, we obtain the force acting on the semiring $$ \boxed{F = 2BI_2R = 2F\frac{I_2R}{Il}} $$ | |||
| #### Answer | |||
| $$F\frac{I_1}{I}\sqrt{1+\frac{L^2}{l^2} + 2\cos\varphi\frac{L}{l}};\quad2F\frac{I_2R}{Il}$$ | |||
| unchanged lines 5 | |||
| ### Statement | ### Statement | ||
| $9.1.3.$ A linear conductor of length $l$, located perpendicular to a magnetic field, is acted upon by a force $F$ if the current in the conductor is $I$. With what force will the magnetic field act on: | $9.1.3.$ A linear conductor of length $l$, located perpendicular to a magnetic field, is acted upon by a force $F$ if the current in the conductor is $I$. With what force will the magnetic field act on: | ||
| @@ -5,7 +5,7 @@Statement | |||
| a) a conductor of length $l + L$ bent at an angle $\varphi$, if the plane of the bend is perpendicular to the magnetic field, and the current in the conductor is $I_1$ | a) a conductor of length $l + L$ bent at an angle $\varphi$, if the plane of the bend is perpendicular to the magnetic field, and the current in the conductor is $I_1$ | ||
| b) a conductor in the form of a semicircle of radius $R$, through which a current $I_2$ flows, if the plane of the semicircle is perpendicular to the magnetic field? | b) a conductor in the form of a semicircle of radius $R$, through which a current $I_2$ flows, if the plane of the semicircle is perpendicular to the magnetic field? | ||
|  | ||
| ### Solution | ### Solution | ||
| a) From Ampere's law, we find the magnetic field induction $B$ $$ F_1 = Il \cdot B \Rightarrow B = \frac{F}{Il}\tag{1} $$ in this case, the numbers acting on the part of the wire with currents $I_1$ and $I_2$, respectively $$ F_1 = I_1l \cdot B; \quad F_2 = I_1L\cdot B $$ We substitute $(1)$: $$ F_1= \frac{I_1}{I}F; \quad F_2 = \frac{I_1L}{Il}F $$ The force acting on a piece of wire bent at an angle $\varphi$ can be found from the cosine law $$ |\vec{F_p}| = |\vec{F_1} + \vec{F_2}| = \sqrt{{F_1}^2+{F_2}^2+2 F_1 F_2\cdot\cos\varphi} $$ $$ \boxed{|\vec{F_p}| = F\frac{I_1}{I}\sqrt{1+\frac{L^2}{l^2} + 2\frac{L}{l}\cdot\cos\varphi}} $$ b) For a conductor made of a semicircle, we will use the same formula for the magnetic field induction $B$ $$ B = \frac{F}{Il} $$ On a small part of a conductor of length $dl$, the Ampere force acts $$ dF = B I_2~dl = B I_2R~d\theta $$ To find the force $F$, we integrate the forces acting on each part of the semiring. $$ F = \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\theta dF = BI_2R\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\theta~d\theta $$ After mathematical transformations, we obtain the force acting on the semiring $$ \boxed{F = 2BI_2R = 2F\frac{I_2R}{Il}} $$ | a) From Ampere's law, we find the magnetic field induction $B$ $$ F_1 = Il \cdot B \Rightarrow B = \frac{F}{Il}\tag{1} $$ in this case, the numbers acting on the part of the wire with currents $I_1$ and $I_2$, respectively $$ F_1 = I_1l \cdot B; \quad F_2 = I_1L\cdot B $$ We substitute $(1)$: $$ F_1= \frac{I_1}{I}F; \quad F_2 = \frac{I_1L}{Il}F $$ The force acting on a piece of wire bent at an angle $\varphi$ can be found from the cosine law $$ |\vec{F_p}| = |\vec{F_1} + \vec{F_2}| = \sqrt{{F_1}^2+{F_2}^2+2 F_1 F_2\cdot\cos\varphi} $$ $$ \boxed{|\vec{F_p}| = F\frac{I_1}{I}\sqrt{1+\frac{L^2}{l^2} + 2\frac{L}{l}\cdot\cos\varphi}} $$ b) For a conductor made of a semicircle, we will use the same formula for the magnetic field induction $B$ $$ B = \frac{F}{Il} $$ On a small part of a conductor of length $dl$, the Ampere force acts $$ dF = B I_2~dl = B I_2R~d\theta $$ To find the force $F$, we integrate the forces acting on each part of the semiring. $$ F = \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\theta dF = BI_2R\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\theta~d\theta $$ After mathematical transformations, we obtain the force acting on the semiring $$ \boxed{F = 2BI_2R = 2F\frac{I_2R}{Il}} $$ | ||
| #### Answer | #### Answer | ||
| $$F\frac{I_1}{I}\sqrt{1+\frac{L^2}{l^2} + 2\cos\varphi\frac{L}{l}};\quad2F\frac{I_2R}{Il}$$ | $$F\frac{I_1}{I}\sqrt{1+\frac{L^2}{l^2} + 2\cos\varphi\frac{L}{l}};\quad2F\frac{I_2R}{Il}$$ | ||
| unchanged lines 5 | |||