Edit to “Solution”
en/9.2.11.md
+1 −1
| ### Statement | |||
| $9.2.11.$ How many times will the magnetic field induction in the center of a ring with current decrease if it is bent at an angle $\alpha$? The current in the ring does not change. | |||
| ### Solution | |||
| @@ -6,8 +6,8 @@Solution | |||
|  | |||
| − | In [9.2.10](../9.2.10) we obtained an expression for the magnetic induction at the center of the ring $$ B_0 = \frac{\mu_0 I}{2R} | ||
| + | In [9.2.10](../9.2.10) we obtained an expression for the magnetic induction at the center of the ring $$ B_0 = \frac{\mu_0 I}{2R}\tag{1} $$ Due to superposition, the magnetic induction at the center of one semicircle will be half of this value: $$ B_1 = B_2 = \frac{B_0}{2} = \frac{\mu_0 I}{4R}\tag{2} $$ From the drawing, it is clear that the angle between the vectors $\vec{B}_1$ and $\vec{B}_2$ is equal to $\alpha$ $$ (\vec{B_1},\vec{B_2})=\alpha $$ The total vector of magnetic induction is equal to the vector sum of the components of magnetic induction of two rings $\vec{B}_1$ and $\vec{B}_2$ $$ \vec{B} = \vec{B_1} + \vec{B_2} $$ We will find the modulus of this vector using the cosine law. $$ B = \sqrt{{B_1}^2 + {B_2}^2 - 2B_1 B_2 \cos\alpha} $$ Taking into account the expression $(2)$, we find the total magnetic induction as $$ B= \frac{\mu_0 I}{4R}\sqrt{2-2\cos\alpha} = \frac{\mu_0 I}{2R}\cdot\sin\frac{\alpha}{2}\tag{3} $$ Dividing the expression $(3)$ by $(1)$, we obtain the desired ratio $$ \boxed{\frac{B}{B_0} = \sin\frac{\alpha}{2}} $$ | ||
| #### Answer | |||
| $$\frac{B}{B_0} =\sin\frac{\alpha}{2}$$ | |||
| ### Statement | ### Statement | ||
| $9.2.11.$ How many times will the magnetic field induction in the center of a ring with current decrease if it is bent at an angle $\alpha$? The current in the ring does not change. | $9.2.11.$ How many times will the magnetic field induction in the center of a ring with current decrease if it is bent at an angle $\alpha$? The current in the ring does not change. | ||
| ### Solution | ### Solution | ||
| @@ -6,8 +6,8 @@Solution | |||
|  |  | ||
| In [9.2.10](../9.2.10) we obtained an expression for the magnetic induction at the center of the ring $$ B_0 = \frac{\mu_0 I}{2R} |
In [9.2.10](../9.2.10) we obtained an expression for the magnetic induction at the center of the ring $$ B_0 = \frac{\mu_0 I}{2R}\tag{1} $$ Due to superposition, the magnetic induction at the center of one semicircle will be half of this value: $$ B_1 = B_2 = \frac{B_0}{2} = \frac{\mu_0 I}{4R}\tag{2} $$ From the drawing, it is clear that the angle between the vectors $\vec{B}_1$ and $\vec{B}_2$ is equal to $\alpha$ $$ (\vec{B_1},\vec{B_2})=\alpha $$ The total vector of magnetic induction is equal to the vector sum of the components of magnetic induction of two rings $\vec{B}_1$ and $\vec{B}_2$ $$ \vec{B} = \vec{B_1} + \vec{B_2} $$ We will find the modulus of this vector using the cosine law. $$ B = \sqrt{{B_1}^2 + {B_2}^2 - 2B_1 B_2 \cos\alpha} $$ Taking into account the expression $(2)$, we find the total magnetic induction as $$ B= \frac{\mu_0 I}{4R}\sqrt{2-2\cos\alpha} = \frac{\mu_0 I}{2R}\cdot\sin\frac{\alpha}{2}\tag{3} $$ Dividing the expression $(3)$ by $(1)$, we obtain the desired ratio $$ \boxed{\frac{B}{B_0} = \sin\frac{\alpha}{2}} $$ | ||
| #### Answer | #### Answer | ||
| $$\frac{B}{B_0} =\sin\frac{\alpha}{2}$$ | $$\frac{B}{B_0} =\sin\frac{\alpha}{2}$$ | ||