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Minchanka [31]
3 years ago
12

When an electric current is flowing through a wire, the force deflecting the charged particles is the greatest when the wire is

what to the magnetic field?
Physics
2 answers:
Makovka662 [10]3 years ago
5 0
The force is greatest when the wire is perpendicular to the magnetic field.

In fact, the expression for the magnetic force is
F=ILB \sin \theta
where I is the current, L the length of the wire, B the magnetic field intensity and \theta is the angle between the directions of I and B.

We can see that F is maximum when \sin \theta=1, so when \theta = 90^{\circ}, so when the wire and the magnetic field are perpendicular to each other.
zzz [600]3 years ago
3 0

Answer:

C.) perpendicular

Explanation:

A particle with an electric charge experiences the maximum deflecting force when it is positioned perpendicular to the magnetic field.

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Answer:

Density of 127 I = \rm 1.79\times 10^{14}\ g/cm^3.

Also, \rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

Explanation:

Given, the radius of a nucleus is given as

\rm r=kA^{1/3}.

where,

  • \rm k = 1.3\times 10^{-13} cm.
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The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.

For the nucleus 127 I,

Mass, M = \rm 2.1\times 10^{-22}\ g.

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.

On comparing with the density of the solid iodine,

\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

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2 years ago
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Answer:

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          0 - 2 s                Constant Velocity           Increasing Velocity

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The position-time graph shares the same spot for two cars.

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