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2 years ago

A long straight wire carries a current of 51.7 A. An electron, traveling at 7.43 × 10^7 m/s, is 5.76 cm from the wire.

Physics

1 answer:

eduard2 years ago

3 0

Answer:

(a) 2.13*10^{-5} N

(b) 2.13*10^{-5} N

Explanation:

the magnitude of the magnetic field generated by the wire is:

$\frac{\mu_0I}{2\pi r}=\frac{(4\pi*10^{-7})(51.7A)}{2\pi(5.76*10^{-2}m)}=1.79*10^{-4}T$

if we assume that the current is in the +y direction, B is in the +z direction.

(a) toward the wire, electron is in the -x direction. The angle between B and v is 90°. By using the following formula we obtain:

$F_1=qvBsin90\°=(1.6*10^{-19}C)(7.43*10^{7}\frac{m}{s})(1.79*10^{-4}T)=2.13*10^{-15}N$

(b) parallel to the wire, electron is in the +y direction. Again angle between B ans v is 90°.

$F_2=qvB=2.13*10^{-15}N$

(c) perpendicular to both previous directions, that is, +z or -z. In this case velocity vector is parallel to the magnetic field vector. Hence:

F3=0N

hope this helps!

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Guys please help me

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Answer:

1) $t=2.26\: s$

2) $S=33.9\: m$

3) $v=26.77\: m/s$

4) $\alpha=55.92$

Explanation:

We can use the following equation:

$y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}$

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

$0=25-0.5*9.81*t^{2}$

$t=2.26\: s$

The equation of the motion in the x-direction is:

$v_{ix}=\frac{S}{t}$

$15=\frac{S}{2.26}$

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$v_{fy}=v_{iy}-gt$

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$v_{fy}=-22.17\: m/s$

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

$v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}$

$v=26.77\: m/s$

The angle of this velocity is:

$tan(\alpha)=\frac{22.17}{15}$

$\alpha=tan^{-1}(\frac{22.17}{15})$

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Then α=55.92° negative from the x-direction.

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2 years ago

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