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FinnZ [79.3K]
3 years ago
8

A closed loop conductor that forms a circle with a radius of 1 m is located in a uniform but changing magnetic field. If the max

imum emf induced in the loop is 5 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
Physics
1 answer:
lora16 [44]3 years ago
8 0

Answer:

1.59 Tesla per second

Explanation:

Given that

Radius of the conductor loop, r = 1 m,

Maximum induced emf to the loop, e = 5 V,

We take an assumption, we assume that the rate which the magnetic field is changing is dB / dt.

We then go ahead to apply Farady's law of electromagnetic induction

e = rate of change of magnetic flux

e = dФ /dt

e = A * dB / dt, on substituting we see that

A = πr²

A = π * 1²

A = π, plug in the value for A

5 = 3.14 * dB / dt

dB / dt = 5/3.142

dB / dt = 1.59 Tesla per second

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Siendo A=pi*r^{2} =pi*(\frac{D}{2} )^{2} =\frac{pi*D^{2} }{4} , donde pi es el número π, r es el radio del conducto y D el diámetro del conducto, entonces:

\frac{pi*D1^{2} }{4}*v1=\frac{pi*D2^{2} }{4}*v2

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  • v1: ?
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Reemplazando:

\frac{pi*(0.06m)^{2} }{4}*v1=\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}

Resolviendo:

v1=\frac{\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}}{\frac{pi*(0.06m)^{2} }{4}}

v1=\frac{(0.04m)^{2} }{(0.06m)^{2}  }*2.6\frac{m}{s}

v1= 1.156 \frac{m}{s}

<u><em>La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 </em></u>\frac{m}{s}<u><em></em></u>

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