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KengaRu [80]
3 years ago
14

"a 2.0-m straight wire carrying a current of 0.60 a is oriented parallel to a uniform magnetic field of 0.50 t. what is the magn

itude of the magnetic force on it? "
Physics
1 answer:
Oksana_A [137]3 years ago
8 0
The magnitude of the magnetic force acting on the wire is zero, because the magnetic field is parallel to the wire.

In fact, the magnetic force exerted by the magnetic field on the wire is
F=ILB \sin \theta
where I is the current in the wire, L the length of the wire, B the magnetic field intensity and \theta the angle between the direction of B and the wire. In our problem, B and the wire are parallel, so the angle is 0^{\circ} and so \sin \theta =0, therefore the magnetic force is zero: F=0.
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GaryK [48]

Answer:

The maximum change in  flux is \Delta \o = 0.1404 \ Wb

The average  induced emf     \epsilon =0.11232 V

Explanation:

   From the question we are told that

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              The distance from the scanner is d = 1.0m

              The  initial magnetic field is  B_i = 0T

               The final magnetic field is B_f = 6.0T

                 The diameter of the loop is  D = 19cm = \frac{19}{100} = 0.19 m

The area of the loop is mathematically represented as

        A  =  \pi [\frac{D}{2} ]^2

             = 3.142 \frac{0.19}{2}

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At maximum the change in magnetic field is mathematically represented as

            \Delta \o = (B_f - B_i)A

  =>      \Delta  \o = (6 -0)(0.02834)

                  \Delta \o = 0.1404 \ Wb

The  average induced emf is mathematically represented as

           \epsilon =  \Delta \o v

              = 0.1404 * 0.80

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Answer:

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Convert 1erg into joule by dimensional method​
Valentin [98]

Answer:

1 * 10^-7 [J]

Explanation:

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