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EastWind [94]
3 years ago
6

A closed loop conductor that forms a circle with a radius of is located in a uniform but changing magnetic field. If the maximum

emf induced in the loop is what is the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies
Physics
1 answer:
kakasveta [241]3 years ago
7 0

Answer:

Explanation:

Complete question:

A closed-loop conductor that forms a circle with a radius of 2.0 m is located in a uniform but changing magnetic field. If the maximum emf induced in the loop is 5.0 V, what is the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies?

A) 0.40T/s

B) 2.5 T/s

C) 0.080 T/s

D) 5.0 T/s

Answer: A) 0.40T/s

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Answer:

<u>Resolving</u><u> </u><u>horizontally</u><u>.</u> :

\sum d_{x} =  - (17.0 \cos 20.0) - (11.0 \cos 35.0) + (30.0 \cos 50.0) + 0 \\  { \underline{d _{x} =  -  5.702 \: m}} \\  \\  \sum d _{y} = (17.0 \sin 20.0) + 12.0 - (11.0  \sin 35.0) - (30.0 \sin 50.0) \\ { \underline{d _{y}  =  - 11.476 \: m}}

therefore, for resultant:

d =  \sqrt{ {d _{y} }^{2} + d _{x}  {}^{2}  }

substitute:

d =  \sqrt{ {( - 5.702)}^{2} +  {( - 11.476)}^{2}  }  \\  \\ d =  \sqrt{164.211}  \\  \\ { \boxed{ \boxed{ \bf{d = 12.8 \: m}}}}

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A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
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Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

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