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kumpel [21]
3 years ago
11

A current I = 17 A is directed along the positive x-axis and perpendicular to a magnetic field. A magnetic force per unit length

of 0.19 N/m acts on the conductor in the negative y-direction. Calculate the magnitude and direction of the magnetic field in the region through which the current passes. magnitude T direction.
Physics
1 answer:
Nonamiya [84]3 years ago
6 0

Answer:

The magnitude of magnetic field is 0.011 T in +z direction.

Explanation:

Given that,

Current, I = 17 A

The magnetic force per unit length, \dfrac{F}{l}=0.19\ N/m

We need to find the magnitude and direction of the magnetic field in the region through which the current passes. The magnetic force acting on the conductor is given by :

F=ilB

B is magnetic field

B=\dfrac{F}{il}\\\\B=\dfrac{0.19}{17}\\\\B=0.011\ T

For direction,

\dfrac{F}{l}(-j)=i(+i) B

The magnetic field is in +z direction. Hence, this is the required solution.

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From the equation, we see that the strength of the field depends on A) the amount of charge that produced the field (Q) and B) the distance from the charge (r), so the correct option is D.</span>
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Explanation:

Given:  uniform electric field E= 82.0 kN/C.

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4 years ago
g 4. A student with a mass of m rides a roller coaster with a loop with a radius of curvature of r. What is the minimum speed th
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where;

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3 years ago
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katrin2010 [14]

Answer:

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Explanation:

The momentum p of an object is the product of its mass m and its velocity v. That is: p = m \cdot v.

Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: v(\text{athlete, before}) = 0 and v(\text{earth, before}) = 0. Therefore:

\begin{aligned}& p(\text{athlete, before}) = 0\end{aligned} and p(\text{earth, before}) = 0.

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Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:

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Therefore:

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\begin{aligned}& m(\text{athlete}) \cdot v(\text{athelete, after}) \\ &= - m(\text{earth}) \cdot v(\text{earth, after})\end{aligned}.

Rewrite this equation to find an expression for v(\text{earth, after}), the speed of the earth after the jump:

\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \end{aligned}.

The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is g = 9.81\; \rm N \cdot kg^{-1}.

\begin{aligned}& m(\text{athlete}) = \frac{664\; \rm N}{9.81\; \rm N \cdot kg^{-1}} \approx 67.686\; \rm N\end{aligned}.

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\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \\ &\approx -2.0 \times 10^{-23}\; \rm m \cdot s^{-1}\end{aligned}.

The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately 2.0 \times 10^{-23}\; \rm m \cdot s^{-1}.

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Remember that an electron has energy levels in an atom or ion, at which each energy level has a specific value.

The energy values will differ from one element to another. So, it can be concluded that each element has a unique pattern of emission lines.

Key terms:        

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²Electronic transition: When an electron passes from one energy level to another, either for the emission or absorption of a photon.

³Ion: An atom electrically charged due to the gain or loss of electrons.

⁴Photon: Elementary particle that constitutes light.

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