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3 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:

eduard3 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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vladimir2022 [97]

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3 years ago

How will the electrostatic force between two electric charges change if the first charge is doubled and the second charge is onl

Vinvika [58]

Answer:

B) $\frac{2}{3}$

Explanation:

The electric force between charges can be determined by;

F = $\frac{kq_{1} q_{2} }{r^{2} }$

Where: F is the force, k is the Coulomb's constant, $q_{1}$ is the value of the first charge, $q_{2}$ is the value of the second charge, r is the distance between the centers of the charges.

Let the original charge be represented by q, so that;

$q_{1}$ = 2q

$q_{2}$ = $\frac{q}{3}$

So that,

F = $q_{1}$ $q_{2}$ x $\frac{k}{r^{2} }$

= 2q x $\frac{q}{3}$ x $\frac{k}{r^{2} }$

= $\frac{2q^{2} }{3}$ x $\frac{k}{r^{2} }$

= $\frac{2}{3}$ x $\frac{kq}{r^{2} }$

F = $\frac{2}{3}$ x $\frac{kq}{r^{2} }$

The electric force between the given charges would change by $\frac{2}{3}$ .

4 0

3 years ago

During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi

geniusboy [140]

Answer:

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

$a_t$ = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

$\omega_i$ = Initial angular velocity = 95 rad/s

$\theta$ = Angle of rotation

$\omega_f$ = Final angular velocity

t = Time taken

Angular acceleration is given by

$\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2$

The angular acceleration is -24.28571 rad/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev$