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Troyanec [42]
2 years ago
9

Determine the velocity of a beam of electrons that goes undeflected when moving perpendicular to an electric and magnetic field.

The electric field and the magnetic field are also perpendicular to each other and have magnitudes 7.7 x 10^3 V/m and 7.5 x 10^-3 T , respectively.
a. What is the radius of the electron orbit if the electric field is turned off?
Physics
1 answer:
maks197457 [2]2 years ago
7 0

Answer:

r = 7.8 × 10 ⁻⁴ m

Explanation:

Given: Electric field E = 7.7 x 10^3 V/m, Magnetic field = 7.5 x 10^-3 T

as |E| = |V| x |B|

⇒  |V| = 7.7 x 10^3 V/m / 7.5 x 10^-3 T = 1.03×10⁶ m/s

Formula for this question is F = qE + qVB  

⇒ F = q VB, ( as Electric component is zero)

the acceleration a = v²/r  and also a = F/m   ( r=radius of curvature) so

so qVB/m = v²/r      

⇒ r = mV/qB

r= (9.11 × 10 ⁻³¹ kg x 1.03×10⁶ m/s) / (1.6 × 10 ⁻¹⁹ C x 7.5 x 10⁻³ T)

r = 7.8 × 10 ⁻⁴ m

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What is the magnitude of your displacement when you follow directions that tell you to walk 225 m in one direction, make a 90° t
Gekata [30.6K]

Start by facing East. Your first displacement is the vector

<em>d</em>₁ = (225 m) <em>i</em>

Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,

<em>d</em>₂ = (350 m) <em>j</em>

Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,

<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )

<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>

The net displacement is

<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃

<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>

and its magnitude is

|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m

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3 years ago
Why do people put rough strips in their bathtubs?
krok68 [10]

Answer:

hi, the reason is for friction so that they don't fall. subscribe to me on you tube. name is my brainly name

Explanation:

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The magnitude of the electric field between two parallel charged plates is 200. An electron moves to the negative plate 5. 0 cm
Mama L [17]

The potential difference between the two ends of the circuit is the electric potential difference. The electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.

<h3>What is an electric field?</h3>

An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.

The given data in the problem is given by;

E is the electric field = (200 N/C)

d is the distance = 5.0 cm.=0.05 m

Q is the charge of electrons= 1.602 x 10^-19 C

The formula for electric potential  is given by;

\rm V=Ed

\rm V=Ed \\\\ \rm V=200 \times 0.05 \\\\ \rm V=  10 \frac{Nm}{C} = 10 \frac{J}{C}  = 10 V.

The work is defined as the product of the potential difference and charge of an electron.

\rm W= 10 \times  1.602 x 10^{-19} \\\\\ \rm W=  1.6 x 10^{-18 }J

Hence the electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.

To learn more about the electric field refer to the link;

brainly.com/question/15071884

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Explanation:

Below is an attachment containing the solution.

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