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Juliette [100K]

4 years ago

15

4.A 31-cm long conducting wire of 9-g carrying 7- A current is placed in a uniform magnetic field. What are the strength and dir

ection of the magnetic field needed to levitate the wire?

1 answer:

Anika [276]4 years ago

7 0

Answer

Given,

Length of the wire,L = 31 cm = 0.31 m

mass of the wire, m = 9 g = 0.009 Kg

Current in the wire,I = 7 A

Magnetic field strength, B= ?

Equating magnetic force to the weight of the wire.

$BIL = m g$

$B=\dfrac{m g}{IL}$

$B=\dfrac{0.009\times 9.81}{7\times 0.31}$

B = 0.0407 T

For Force to be upward magnetic field direction should be outward of the plane of paper.

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Zanzabum

Answer:

The current needed to transmit Power of 4 W is 28.47 A

Solution:

As per the question:

Length of the antenna, $L_{a} = 1 m$

Frequency, $\vartheta = 1.5 MHz = 1.5\times 10^{6} Hz$

Power transmitted, $P_{t} = 4 W$

Now,

For a monopole antenna:

$\lambda_{a} = \frac{c}{\vartheta}$

where

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c = speed of light in vacuum

$\lambda_{a} = \frac{3\times 10^{8}}{1.5\times 10^{6}} = 200 m$

Now,

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The resistance is calculated as:

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$R = \frac{1}{2}(\frac{1}{200)^{2}\times 80\pi^{2} = 9.869\times 10^{- 3} = 9.869 m\Omega$

$P_{radiated} = P_{t}$

$P_{radiated} = \frac{R}{I^{2}}$

Now, the current I is given by:

$I = \sqrt{\frac{2P_{t}}{R}} = \sqrt{\frac{2\times 4}{9.869\times 10^{- 3}}} = 28.47 A$

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3 years ago

The potential difference between the plates of a parallel plate capacitor is 35 V and the electric field between the plates has

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Answer:

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4 years ago