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Mrac [35]
3 years ago
14

A wire has a current that flows to the right. Which describes the magnetic field produced by the current? straight lines along t

he wire to the right moving the same direction of the current straight lines along the wire to the left moving the opposite direction of the current circles around the wire that go in at the top of the screen and moves in a counterclockwise direction circles around the wire that go in at the bottom of the screen and moves in a clockwise direction
Physics
2 answers:
Flura [38]3 years ago
7 0

Answer:

D.) circles around the wire that go in at the bottom of the screen and moves in a clockwise direction

Explanation:

It is correct on Edge

kirill115 [55]3 years ago
4 0

Answer:

the correct ones are: circles that come out of the upper part of the screen and enter the lower part

Explanation:

When an electric current flows through a wire, a magnetic field is produced, the direction of which is given by the right-hand rule, the thumb goes in the direction of the current, and the other closed fingers indicate the direction of the magnetic field.

in that case the thumb points to the right, which is why the fingers are outgoing and incoming circles on the pages, therefore the magnetic field creates concentric circles around the wire.

With these explanations we review the different statements,

the correct ones are: circles that come out of the upper part of the screen and enter the lower part,

   

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Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

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4 years ago
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nikdorinn [45]

Answer:

Explanation:

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