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Oksana_A [137]
3 years ago
9

A 1.0-m long wire is carrying a certain amount of current. The wire is placed perpendicular to a magnetic field of strength 0.20

T. If the wire experiences a force of 0.60 N, what is the magnitude of the current moving through the wire?
Physics
1 answer:
Marysya12 [62]3 years ago
5 0

For a current-carrying wire running perpendicular to a magnetic field, the magnetic force acting on the wire is given by:

F = ILB

F = magnetic force, I = current, L = wire length, B = magnetic field strength

Given values:

F = 0.60N, L = 1.0m, B = 0.20T

Plug in and solve for I:

0.60 = I(1.0)(0.20)

I = 3.0A

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A rock is suspended by a light string. When the rock is in air, the tension in the string is 51.9 N . When the rock is totally i
Luden [163]

Answer:

\rho _{liquid}=1995.07kg/m^{3}

Explanation:

When the rock is immersed in unknown liquid the forces that act on it are shown as under

1) Tension T by the string

2) Weight W of the rock

3) Force of buoyancy due to displaced liquid B

For equilibrium we have T_{3}+B = W_{rock}

T_{3}+\rho _{Liquid}V_{rock}g=W_{rock}.....(\alpha)

When the rock is suspended in air for equilibrium we have

T_{1}=W_{rock}....(\beta)

When the rock is suspended in water for equilibrium we have

T_{2} + \rho _{water}V_{rock}g=W_{rock}.....(\gamma)

Using the given values of tension and solving α,β,γ simultaneously for \rho _{Liquid} we get

W_{rock}=51.9N\\31.6+1000\times V_{rock}\times g=51.9N\\\\11.4+\rho _{liquid}V_{rock}g=51.9N\\\\

Solving for density of liquid we get

\rho _{liquid}=\frac{51.9-11.4}{51.9-31.6}\times 1000

\rho _{liquid}=1995.07kg/m^{3}

5 0
2 years ago
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