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en/9.1.5.md
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| ### Statement | |||
| − | $9.1.5.$ A conductor of mass $0.16 | ||
| + | $9.1.5.$ A conductor of mass $0.16\mathrm{~kg}$ and length $80\mathrm{~cm}$ is suspended horizontally in a vertical uniform magnetic field on two thin threads. The ends of the conductor are connected to a current source by flexible wires located outside the field. Find the angle by which the suspension threads deviate from the vertical if a current of $2\mathrm{~A}$ flows through the conductor and the magnetic field induction is $1\mathrm{~T}$. | ||
| ### Solution | |||
|  | |||
| Ampere force acting on a conductor $$ F = Il\cdot B $$ From Newton's second law, the condition of equilibrium on the horizontal and vertical axes $$ T\cos\alpha = mg;\quad T\sin\alpha = F $$ dividing one expression by the other, we get $$ \tan\alpha = \frac{F}{mg} $$ $$ \boxed{\alpha = \arctan \left(\frac{IlB}{mg}\right) = 45^{\circ}} $$ | |||
| #### Answer: $\alpha = 45^{\circ}$ | |||
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| ### Statement | ### Statement | ||
| $9.1.5.$ A conductor of mass $0.16 |
$9.1.5.$ A conductor of mass $0.16\mathrm{~kg}$ and length $80\mathrm{~cm}$ is suspended horizontally in a vertical uniform magnetic field on two thin threads. The ends of the conductor are connected to a current source by flexible wires located outside the field. Find the angle by which the suspension threads deviate from the vertical if a current of $2\mathrm{~A}$ flows through the conductor and the magnetic field induction is $1\mathrm{~T}$. | ||
| ### Solution | ### Solution | ||
|  |  | ||
| Ampere force acting on a conductor $$ F = Il\cdot B $$ From Newton's second law, the condition of equilibrium on the horizontal and vertical axes $$ T\cos\alpha = mg;\quad T\sin\alpha = F $$ dividing one expression by the other, we get $$ \tan\alpha = \frac{F}{mg} $$ $$ \boxed{\alpha = \arctan \left(\frac{IlB}{mg}\right) = 45^{\circ}} $$ | Ampere force acting on a conductor $$ F = Il\cdot B $$ From Newton's second law, the condition of equilibrium on the horizontal and vertical axes $$ T\cos\alpha = mg;\quad T\sin\alpha = F $$ dividing one expression by the other, we get $$ \tan\alpha = \frac{F}{mg} $$ $$ \boxed{\alpha = \arctan \left(\frac{IlB}{mg}\right) = 45^{\circ}} $$ | ||
| #### Answer: $\alpha = 45^{\circ}$ | #### Answer: $\alpha = 45^{\circ}$ | ||
| unchanged lines 5 | |||