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| ### Statement |
| ### Statement |
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| $11.2.3.$ [Insert the problem statement] |
| $11.2.3.$ |
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| A conducting ring having a jumper with an electric light bulb in diameter is |
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| moved in the magnetic field of the solenoid with current so that the plane of |
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| the ring is perpendicular to the axis of the solenoid, and the jumper with the |
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| light bulb is perpendicular to the direction of the speed of movement of the |
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| ring. In ring positions A and B, the light bulb glows, and in position C, it goes |
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| out. Explain the observed effect. |
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| ### Solution |
| ### Solution |
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| A long solenoid with current produces a magnetic field with the following characteristics: |
| A long solenoid with current produces a magnetic field with the following characteristics: |
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| · Inside, far from the ends, the field is uniform and parallel to the axis (position C). | | · Inside, far from the ends, the field is uniform and parallel to the axis (position C). |
| · Near the ends (positions A and B), the field is non‑uniform: the field lines spread out and there is both a radial and an axial gradient. The intensity decreases as one moves away from the center. | | · Near the ends (positions A and B), the field is non‑uniform: the field lines spread out and there is both a radial and an axial gradient. The intensity decreases as one moves away from the center. |
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| Taking this into account, let us analyze the motion of the ring and the magnetic flux. | | Taking this into account, let us analyze the motion of the ring and the magnetic flux. |
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| The ring moves horizontally, keeping its plane perpendicular to the axis of the solenoid. | | The ring moves horizontally, keeping its plane perpendicular to the axis of the solenoid. |
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| · At A and B (ends): when the ring moves, it passes through regions where the field intensity varies with position. Therefore, the magnetic flux through the area of the ring changes with time. | | · At A and B (ends): when the ring moves, it passes through regions where the field intensity varies with position. Therefore, the magnetic flux through the area of the ring changes with time. |
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| At C (center): the field is uniform. When shifting the ring laterally, the value of$ \mathbf{B}$ at every point on its surface is the same, and the flux then remains constant. | | At C (center): the field is uniform. When shifting the ring laterally, the value of$ \mathbf{B}$ at every point on its surface is the same, and the flux then remains constant. |
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| We apply Faraday's Law. | | We apply Faraday's Law. |
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| The electromotive force induced in the closed circuit (ring + bridge) is: | | The electromotive force induced in the closed circuit (ring + bridge) is: |
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| $\mathcal{E} = -\frac{d\Phi}{dt}$ | | $\mathcal{E} = -\frac{d\Phi}{dt}$ |
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| · At A and B | | · At A and B |
| $ d\Phi/dt \neq 0 $ → an emf is induced → current flows through the bulb → it lights up. | | $ d\Phi/dt \neq 0 $ → an emf is induced → current flows through the bulb → it lights up. |
| · At C | | · At C |
