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Nostrana [21]
3 years ago
13

Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside thi

s metal resistor?
Physics
1 answer:
Zanzabum3 years ago
7 0

Complete question:

Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.

Answer:

The electric field inside this metal resistor is 3125 V/m

Explanation:

Given;

length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m

diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m

the potential difference due to electric field between the two ends of the resistor, V = 10 V

The electric field inside this metal resistor is given by;

ΔV = EL

where;

ΔV is change in electric potential

E = ΔV / L

E = 10 / (3.2 x 10⁻³ )

E = 3125 V/m

Therefore, the electric field inside this metal resistor is 3125 V/m

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Given Information:

Radius = r = 0.5 m

Magnetic field = 1.0 T

Required Information:

Period = T = ?

Speed = v = ?

Kinetic energy = KE = ?

Answer:

Period = 0.13x10⁻⁶ seconds

speed = 24.16x10⁶ m/s

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Explanation:

(a) period

The time period of alpha particle is related to its orbital speed as

T = 2πr/v  eq. 1

According to newton's law

F = ma

Force due to magnetic field is given by

F = qvB

qvB = ma

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qB = mv/r

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substitute the eq. 2 in eq. 1

T = 2πr/qBr/m

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T = 2π/qB/m

T = 2πm/qB

T = 2π*6.65x10⁻²⁷/2*1.602x10⁻¹⁹*1

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(b) speed

From equation 1

T = 2πr/v

v = 2πr/T

v = 2π*0.5/0.13x10⁻⁶

v = 24.16x10⁶ m/s

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Kinetic energy is given by

KE = 0.5mv²

KE = 0.5*6.65x10⁻²⁷*(24.16x10⁶)²

KE = 1.94x10⁻¹² J

since 1 electron volt has 1.602x10⁻¹⁹ J

KE = 1.94x10⁻¹²/1.602x10⁻¹⁹

KE = 12.11 MeV

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