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Cerrena [4.2K]
3 years ago
11

A 170.4-turn circular loop of radius 0.08 meters with a magnetic field B passing through the area bounded by the loop at right a

ngles has an Emf of 0.050 Volts induced around the loop by the rising magnetic field. How fast is the magnetic field rising in tesla per second
Physics
1 answer:
jolli1 [7]3 years ago
8 0

Answer:

The speed of the magnetic field rising is   \frac{\Delta B}{\Delta t}  =0.0147 \ T/s

Explanation:

Mathematically Induced Emf is as follows

             e = - \frac{\Delta B}{\Delta t} *N * Area

The objective of this solution is to obtain speed in Tesla per second which the same as \frac{\Delta B}{\Delta t} since the unit of magnetic field B is Tesla and Time is seconds

             So making  \frac{\Delta B}{\Delta t}  the subject

                          \frac{\Delta B}{\Delta t }  =\frac{Emf}{N*Area}

The \ Area = \pi r^2

               = 3.142 * (0.08)^2

              =0.020m

N is given as 170.4 turns

         Therefore  \  \frac{\Delta B}{\Delta t} = \frac{0.050}{170.4*0.02}

                                  =0.0147 \ T/s

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Answer:

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i = 12.5 * 22 / (12.5 - 22) = -28.9 cm

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An ore sample weighs 17.50 N in air. When the sample
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Answer with Explanation:

We are given that

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We have to find the total volume and the density of the sample.

We know that

Tension, T=W-F_b

F_b=buoyancy force

T=Tension force

W=Weight

By using the formula

11.2=17.5-F_b

F_b=17.5-11.2=6.3 N

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Substitute the values then we get

6.3=9.8\times 1000\times V_{object}

V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3

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Substitute the values then, we get

\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3

Density of the sample=2.78\times 10^{3} kgm^{-3}

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