Правка разделов «Statement», «Solution», «Answer»
en/11.1.9.md
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| @@ -1,24 +1,14 @@ | |||
| ### Statement | |||
| − | $11.1.9.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| $11.1.9.$ A square closed loop of wire with side length $b$ and resistance per unit length $\rho$ moves with a constant velocity $v$ through the gap of an electromagnet. The magnetic field in the gap is uniform, and its induction is equal to $B$. Assuming the field outside the gap is zero, determine the energy converted into heat for the cases when the width of the gap $a$ in the direction of the loop's motion is less than $b$ and greater than $b$, and in the perpendicular direction is greater than $b$. | |||
| − | <div align="center"> | ||
| − | <img src="https://savchenkosolutions.com/11/11.1.9/statement.png" alt="For problem 11.1.9"> | ||
| − | </div> | ||
| + |  | ||
| ### Solution | |||
| − | <div align="center"> | ||
| − | <img src="https://savchenkosolutions.com/11/11.1.9/drawing.png" alt="Loop motion"> | ||
| − | </div> | ||
| + |  | ||
| − | <b>1. General principles</b> | ||
| + | <b>1. General principles</b>\ | ||
| The total length of the wire of the square loop is $4b$. Therefore, the total electrical resistance of the loop is: | |||
| $$R = 4b\rho$$ | |||
| When any vertical side of the frame crosses the boundary of the magnetic field (one side is inside the field and the opposite one is outside), the magnetic flux through the frame changes. An induced electromotive force (EMF) arises in the loop: | |||
| $$\varepsilon = \left| \frac{d\Phi}{dt} \right| = Bbv$$ | |||
| According to Ohm's law, the resulting induction current flows through the <i>entire</i> closed loop: | |||
| $$I = \frac{\varepsilon}{R} = \frac{Bbv}{4b\rho} = \frac{Bv}{4\rho}$$ | |||
| @@ -31,7 +21,7 @@Solution | |||
| The thermal power dissipated in the <i>entire</i> frame during the change of flux is: | |||
| $$P = I^2 R = \left(\frac{Bv}{4\rho}\right)^2 \cdot 4b\rho = \frac{B^2 v^2 b}{4\rho}$$ | |||
| − | <b>2. Case $a > b$ (the gap is wider than the frame)</b> | ||
| + | <b>2. Case $a > b$ (the gap is wider than the frame)</b>\ | ||
| The magnetic flux changes, and the current flows only during two stages of motion: | |||
| 1. <b>Entering the field:</b> the front side of the frame enters the gap. This lasts until the frame completely enters the field. The duration of this stage is $t_1 = \frac{b}{v}$. | |||
| 2. <b>Exiting the field:</b> the front side leaves the gap, while the back side is still inside. The duration of this stage is $t_2 = \frac{b}{v}$. | |||
| While the frame moves entirely within the field, the magnetic flux is constant, there is no current, and no heat is dissipated. | |||
| The total time of heat dissipation is $t = t_1 + t_2 = \frac{2b}{v}$. | |||
| @@ -41,7 +31,7 @@Solution | |||
| The total dissipated energy is: | |||
| $$W_1 = P \cdot t = \frac{B^2 v^2 b}{4\rho} \cdot \frac{2b}{v} = \frac{B^2 v b^2}{2\rho}$$ | |||
| − | <b>3. Case $a < b$ (the gap is narrower than the frame)</b> | ||
| + | <b>3. Case $a < b$ (the gap is narrower than the frame)</b>\ | ||
| Here, the stages of heat dissipation are different: | |||
| 1. <b>Front side passage:</b> the front side of the frame passes through the gap from beginning to end. The duration of this stage is $t_1 = \frac{a}{v}$. | |||
| 2. <b>Back side passage:</b> the back side of the frame passes through the gap. The duration of this stage is $t_2 = \frac{a}{v}$. | |||
| When the gap is located between the front and back sides (both vertical sides are outside the field), the flux through the frame is constant and equals $Bab$. There is no current at this moment. | |||
| The total time of heat dissipation is $t = t_1 + t_2 = \frac{2a}{v}$. | |||
| The total dissipated energy is: | |||
| $$W_2 = P \cdot t = \frac{B^2 v^2 b}{4\rho} \cdot \frac{2a}{v} = \frac{B^2 v a b}{2\rho}$$ | |||
| @@ -54,7 +44,3 @@Solution | |||
| #### Answer | |||
| $$W = \frac{B^2 v b^2}{2\rho} \text{ for } a > b$$ | |||
| $$W = \frac{B^2 v a b}{2\rho} \text{ for } a < b$$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,24 +1,14 @@ | |||
| ### Statement | ### Statement | ||
| $11.1.9.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $11.1.9.$ A square closed loop of wire with side length $b$ and resistance per unit length $\rho$ moves with a constant velocity $v$ through the gap of an electromagnet. The magnetic field in the gap is uniform, and its induction is equal to $B$. Assuming the field outside the gap is zero, determine the energy converted into heat for the cases when the width of the gap $a$ in the direction of the loop's motion is less than $b$ and greater than $b$, and in the perpendicular direction is greater than $b$. | $11.1.9.$ A square closed loop of wire with side length $b$ and resistance per unit length $\rho$ moves with a constant velocity $v$ through the gap of an electromagnet. The magnetic field in the gap is uniform, and its induction is equal to $B$. Assuming the field outside the gap is zero, determine the energy converted into heat for the cases when the width of the gap $a$ in the direction of the loop's motion is less than $b$ and greater than $b$, and in the perpendicular direction is greater than $b$. | ||
| <div align="center"> |  | ||
| <img src="https://savchenkosolutions.com/11/11.1.9/statement.png" alt="For problem 11.1.9"> | |||
| </div> | |||
| ### Solution | ### Solution | ||
| <div align="center"> |  | ||
| <img src="https://savchenkosolutions.com/11/11.1.9/drawing.png" alt="Loop motion"> | |||
| </div> | |||
| <b>1. General principles</b> | <b>1. General principles</b>\ | ||
| The total length of the wire of the square loop is $4b$. Therefore, the total electrical resistance of the loop is: | The total length of the wire of the square loop is $4b$. Therefore, the total electrical resistance of the loop is: | ||
| $$R = 4b\rho$$ | $$R = 4b\rho$$ | ||
| When any vertical side of the frame crosses the boundary of the magnetic field (one side is inside the field and the opposite one is outside), the magnetic flux through the frame changes. An induced electromotive force (EMF) arises in the loop: | When any vertical side of the frame crosses the boundary of the magnetic field (one side is inside the field and the opposite one is outside), the magnetic flux through the frame changes. An induced electromotive force (EMF) arises in the loop: | ||
| $$\varepsilon = \left| \frac{d\Phi}{dt} \right| = Bbv$$ | $$\varepsilon = \left| \frac{d\Phi}{dt} \right| = Bbv$$ | ||
| According to Ohm's law, the resulting induction current flows through the <i>entire</i> closed loop: | According to Ohm's law, the resulting induction current flows through the <i>entire</i> closed loop: | ||
| $$I = \frac{\varepsilon}{R} = \frac{Bbv}{4b\rho} = \frac{Bv}{4\rho}$$ | $$I = \frac{\varepsilon}{R} = \frac{Bbv}{4b\rho} = \frac{Bv}{4\rho}$$ | ||
| @@ -31,7 +21,7 @@Solution | |||
| The thermal power dissipated in the <i>entire</i> frame during the change of flux is: | The thermal power dissipated in the <i>entire</i> frame during the change of flux is: | ||
| $$P = I^2 R = \left(\frac{Bv}{4\rho}\right)^2 \cdot 4b\rho = \frac{B^2 v^2 b}{4\rho}$$ | $$P = I^2 R = \left(\frac{Bv}{4\rho}\right)^2 \cdot 4b\rho = \frac{B^2 v^2 b}{4\rho}$$ | ||
| <b>2. Case $a > b$ (the gap is wider than the frame)</b> | <b>2. Case $a > b$ (the gap is wider than the frame)</b>\ | ||
| The magnetic flux changes, and the current flows only during two stages of motion: | The magnetic flux changes, and the current flows only during two stages of motion: | ||
| 1. <b>Entering the field:</b> the front side of the frame enters the gap. This lasts until the frame completely enters the field. The duration of this stage is $t_1 = \frac{b}{v}$. | 1. <b>Entering the field:</b> the front side of the frame enters the gap. This lasts until the frame completely enters the field. The duration of this stage is $t_1 = \frac{b}{v}$. | ||
| 2. <b>Exiting the field:</b> the front side leaves the gap, while the back side is still inside. The duration of this stage is $t_2 = \frac{b}{v}$. | 2. <b>Exiting the field:</b> the front side leaves the gap, while the back side is still inside. The duration of this stage is $t_2 = \frac{b}{v}$. | ||
| While the frame moves entirely within the field, the magnetic flux is constant, there is no current, and no heat is dissipated. | While the frame moves entirely within the field, the magnetic flux is constant, there is no current, and no heat is dissipated. | ||
| The total time of heat dissipation is $t = t_1 + t_2 = \frac{2b}{v}$. | The total time of heat dissipation is $t = t_1 + t_2 = \frac{2b}{v}$. | ||
| @@ -41,7 +31,7 @@Solution | |||
| The total dissipated energy is: | The total dissipated energy is: | ||
| $$W_1 = P \cdot t = \frac{B^2 v^2 b}{4\rho} \cdot \frac{2b}{v} = \frac{B^2 v b^2}{2\rho}$$ | $$W_1 = P \cdot t = \frac{B^2 v^2 b}{4\rho} \cdot \frac{2b}{v} = \frac{B^2 v b^2}{2\rho}$$ | ||
| <b>3. Case $a < b$ (the gap is narrower than the frame)</b> | <b>3. Case $a < b$ (the gap is narrower than the frame)</b>\ | ||
| Here, the stages of heat dissipation are different: | Here, the stages of heat dissipation are different: | ||
| 1. <b>Front side passage:</b> the front side of the frame passes through the gap from beginning to end. The duration of this stage is $t_1 = \frac{a}{v}$. | 1. <b>Front side passage:</b> the front side of the frame passes through the gap from beginning to end. The duration of this stage is $t_1 = \frac{a}{v}$. | ||
| 2. <b>Back side passage:</b> the back side of the frame passes through the gap. The duration of this stage is $t_2 = \frac{a}{v}$. | 2. <b>Back side passage:</b> the back side of the frame passes through the gap. The duration of this stage is $t_2 = \frac{a}{v}$. | ||
| When the gap is located between the front and back sides (both vertical sides are outside the field), the flux through the frame is constant and equals $Bab$. There is no current at this moment. | When the gap is located between the front and back sides (both vertical sides are outside the field), the flux through the frame is constant and equals $Bab$. There is no current at this moment. | ||
| The total time of heat dissipation is $t = t_1 + t_2 = \frac{2a}{v}$. | The total time of heat dissipation is $t = t_1 + t_2 = \frac{2a}{v}$. | ||
| The total dissipated energy is: | The total dissipated energy is: | ||
| $$W_2 = P \cdot t = \frac{B^2 v^2 b}{4\rho} \cdot \frac{2a}{v} = \frac{B^2 v a b}{2\rho}$$ | $$W_2 = P \cdot t = \frac{B^2 v^2 b}{4\rho} \cdot \frac{2a}{v} = \frac{B^2 v a b}{2\rho}$$ | ||
| @@ -54,7 +44,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $$W = \frac{B^2 v b^2}{2\rho} \text{ for } a > b$$ | $$W = \frac{B^2 v b^2}{2\rho} \text{ for } a > b$$ | ||
| $$W = \frac{B^2 v a b}{2\rho} \text{ for } a < b$$ | $$W = \frac{B^2 v a b}{2\rho} \text{ for } a < b$$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||