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Consider a wire of length L = 0.30m that runs north-south on a horizontal surface. There is a current of I = 0.50A flowing north in the wire. (The rest of the circuit, which actually delivers this current, is not shown.)

The Earth's magnetic field at this location has a magnitude of 0.50 (or, in SI units, 0.5 X10^-4 Tesla) and points north and 38 degrees down from the horizontal, toward the ground. What is the size of the magnetic force on the wire due to the Earth's magnetic field? In considering the agreement of units, recall that 1T=1N/(AxM).

a. Express your answer in newtons to two significant figures.

b. Now assume that a strong, uniform magnetic field of size 0.55T pointing straight down is applied. What is the size of the magnetic force on the wire due to this applied magnetic field? Ignore the effect of the Earth's magnetic field.

Respuesta :

Answer:

Part a)

[tex]F = 4.6 \times 10^{-6} N[/tex]

Part b)

[tex]F = 8.2 \times 10^{-2} N[/tex]

Explanation:

Part a)

As we know that the magnetic force on the current carrying wire is given as

[tex]F = i(\vec L \times \vec B)[/tex]

so we have

[tex]F = iLB sin\theta[/tex]

[tex]F = (0.50)(0.30)(0.5 \times 10^{-4}) sin38[/tex]

so we have

[tex]F = 4.6 \times 10^{-6} N[/tex]

Part b)

Now magnetic field is changed to 0.55 T

so we will have

[tex]F = (0.50)(0.30)(0.55) sin90[/tex]

[tex]F = 8.2 \times 10^{-2} N[/tex]

a. The magnetic force should be  F = 4.6*10^-6N

b. The magnetic force should be F = 8.2*10^-2N

Calculation of the magnetic force:

a. Here the magnetic force should be

= (0.50)(0.30)(0.50*10^-4) sin38

= 4.6*10^-6N

b. Now the magnetic force in the case when the magnetic field is changed to 0.55 T

So, the magnetic force is

= (0.50)(0.30)(0.55)sin 90

= 8.2*10^-2N

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