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galina1969 [7]
3 years ago
12

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu

late the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.
Physics
1 answer:
Alekssandra [29.7K]3 years ago
6 0

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

F = BIl \ sin(\theta)

(a) When the angle, θ = 60 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N

(b) When the angle, θ = 90 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N

(c) When the angle, θ = 120 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N

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\:  \underline{ \boxed{  { \color{gray}\frak{\huge \: Answer : }}}}

<h3> Collector Current at Saturation :</h3>

  • \sf \large {I_{C(SAT)} = \frac{ V_{CC} }{R_C} }

\:  \:

  • \sf \large {I_{C(SAT)} = \frac{12}{2.2 \times 10 ^{3} } }

\:  \:

  • \bold{ \bf \large {I_{C(SAT)} =5.45mA }}

\:  \:

________________________________

<h3> Value Of Cut - off Voltage : </h3>

  • \sf \large \: V_{CS(cut-off)} = V_{CC}

\:  \:

Therefore ,

\:  \:

  • \bf \large \: V_{CS(cut-off)} = \: 12v

\:  \:

________________________________

{ \bf\large \: Base \:  Current , I_B =  \frac{V_{CC}}{R_C} }

  • \sf \large \: I_B ={ \frac{12}{2.2 \times 10 ^{3} } }

\:  \:

  • \bf \large \: I_B = 50μA

\:  \:

_______________________________

<h3>Collector Current ,</h3>

  • \sf \large \: I_C = β * I_B

\:  \:

  • \sf \large \: I_C = 50 * 50 * 10^{-6}

\:  \:

  • \bf \large \: I_C = 2.5mA

\:  \:

________________________________

<h3> Collector to emitter Voltage : </h3>

  • \sf \large \: V_{CE} = V_{CC} - ( I_C * R_C )

\:  \:  \:

  • \sf \large \: V_{CE} = 12 - ( 2.5 * 10-³ * 2.2 * 10³ )

\:  \:

  • \bf \large \: V_{CE} = 6.5v

________________________________

<h3>Q - point are :</h3>

  • \sf \large \: I_{CEQ} = 2.5mA

\:  \:

  • \sf \large \: V_{CEQ} = 6.5v

________________________________

<h3>Q - point located on the DC load line as shown in fig ~</h3><h3 /><h3 /><h3 /><h3 />

________________________________

Hope Helps!:)

\:  \:  \:  \:

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