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timurjin [86]
3 years ago
10

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 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:
galina1969 [7]3 years ago
3 0

Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a)60 (b)90 (c)120

Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

F=I \times B \times L \times sin(\theta)

So from data

F = 5.20 A \times 0.430 T \times 2.80 sin(\theta)\\\\F=6.2608 sin(\theta) N

Now sub parts

(a)

\theta=60^{o}\\\\Force = 6.2608 sin(60^{o}) N\\\\Force = 5.42 N

(b)

\theta=90^{o}\\\\Force = 6.2608 sin(90^{o}) N\\\\Force = 6.26 N

(c)

\theta=120^{o}\\\\Force = 6.2608 sin(120^{o}) N\\\\Force = 5.42 N

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We have that the Average Force is given

Avg F=2.0488N

 

From the question we are told

Masses=

0.10

0.25

0.40

0.50

0.60

0.75

1

1.5

2

Accelerations

9.5

4.2

2.3

2.1

1.7

1.3

1.1

0.7

0.5

Generally the equation for the Mean is mathematically given as

For Mass

X=\frac{\sum x}{n}\\\\X=\frac{0.10 +0.25+0.40+0.50+0.60+0.75+1+1.5+2}{9}\\\\X=0.788g

For Acceleration

X'=\frac{\sum x}{n}\\\\X'=\frac{9.5+4.2+2.3+2.1+1.7+1.3+1.1+0.7+0.5}{9}\\\\X'=2.6m/s^2

Since Average Acceleration and mass are found above

Therefore

The Average Force is given as

Generally the equation for the Average Force is mathematically given as

Avg F=Avg m *Avg a\\\\Avg F=0.788*2.6\\\\Avg F=2.0488

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