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2 years ago

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The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the

wire?

1 answer:

FrozenT [24]2 years ago

4 0

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

$B=\frac{2\pi*a }{u*I}$

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

$I=\frac{2\pi*a*B }{u}$ Formula(2)

$B=0.2*10^{-5} T = 0.2*10^{-5} \frac{weber}{m^{2} }$

a =8cm=0.08m

$u=4*\pi *10^{-7} \frac{Weber}{A*m}$

We replace the known information in the formula (2)

$I=\frac{2\pi*0.08*0.2*10^{-5} }{4\pi *10x^{-7} }$

I=0.8 A

Answer: The electric current in the wire is 0.8 A

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Answer:

Speed of larger piece is $\dfrac{V\sqrt{2}}{3}$

Explanation:

We apply the principle of conservation of momentum.

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After the collision,

Vertical momentum = momentum of piece in y-direction + y-component of momentum of larger piece = $mV + 3mv_{ly}$

Here, $v_{ly}$ is the y-component of velocity of larger piece.

This is equal to 0, since the initial momentum is 0.

$v_{ly}=\dfrac{V}{3}$

Horizontal momentum = momentum of piece in x-direction + x-component of momentum of larger piece = $mV + 3mv_{lx}$

Here, $v_{lx}$ is the x-component of velocity of larger piece.

This is also equal to 0, since the initial momentum is 0.

$v_{lx}=\dfrac{V}{3}$

The velocity of the larger piece, $v_l$ , is the resultant of $v_{lx}$ and $v_{ly}$ . Since they are mutually perpendicular,

$v_l = \sqrt{v_{ly}^2+v_{lx}^2}= \sqrt{\left(\dfrac{V}{3}\right)^2+\left(\dfrac{V}{3}\right)^2}$

$v_l = \dfrac{V\sqrt{2}}{3}$

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2 years ago

A positive point charge (q = +5.45 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 1.49 m. A p

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Answer:

$r_{B} = 3.34\ m$

Solution:

As per the question:

Point charge, q = $5.45\times 10^{- 8}\ C$

Test charge, $q_{o} = 4.96\times 10^{- 11}\ C$

Work done by the electric force, $W_{AB} = - 9.05\times 10^{- 9}\ J$

Now,

We know that the electric potential at a point is given by:

$V = \frac{kqq'}{r}$

where

r = separation distance between the charges.

Also,

The work done by the electric force i moving a test charge from point A to B in an electric field:

$W_{AB} = kqq_{o}(\frac{1}{r_{B} - \frac{1}{r_{A}}}$

$- 9.05\times 10^{- 9} = 9\times 10^{9}\times 5.45\times 10^{- 8}\times 4.96\times 10^{- 11}(\frac{1}{r_{B}} - \frac{1}{1.49}}$

$- 0.3719 = (\frac{1}{r_{B}} - 0.67}$

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Answer:

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Answer:

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Explanation:

Considering the complete question attached in figure below.

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As dimensions of new balance wheel are one-third of their original values

$R_{new}=\frac{R}{3}$

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