Question1. Copper metal wire has a linear density of $0.010 \mathrm{~kg} / \mathrm{m}$ . A sample of this wire is stretched horizontally in an area where the horizontal component of Earth's magnetic field of strength 0.000020 T passes through the wire at right angles. a) What current must be applied to the wire if the weight of the entire wire is supported by the mangetic force? (3 marks) b) If this current is applied, what might happen to the wire? (1 mark)

Studdy Solution

STEP 1

What is this asking? How much electrical current do we need to make the magnetic force on a horizontal wire equal to the wire's weight, and what happens to the wire with that current? Watch out! Don't mix up the magnetic field strength and the magnetic force!
Also, make sure all your units are compatible – we're working with kilograms, meters, and Teslas!

STEP 2

1. Formula for Magnetic Force on a Wire
2. Calculate the Current

STEP 3

Alright, let's start with the formula for the magnetic force on a current-carrying wire: $F = I \cdot L \cdot B$ , where $F$ is the magnetic force, $I$ is the current, $L$ is the length of the wire, and $B$ is the magnetic field strength.
This tells us how much force the magnetic field exerts on the wire.

STEP 4

We want the magnetic force ( $F$ ) to be equal to the wire's weight, which is just mass ( $m$ ) times gravity ( $g$ ).
So, we have $F = m \cdot g$ .
Gravity is approximately $9.8 \, \text{m/s}^2$ .

STEP 5

We're given the linear density ( $\mu$ ), which is the mass per unit length: $\mu = \frac{m}{L}$ .
We can rewrite this as $m = \mu \cdot L$ .

STEP 6

Now, we can substitute this expression for mass into our weight equation: $F = (\mu \cdot L) \cdot g$ .
See how the length ( $L$ ) is now involved?
This is great because it also appears in the magnetic force formula!

STEP 7

Now, we can set the magnetic force equal to the weight: $I \cdot L \cdot B = \mu \cdot L \cdot g$ .
Our goal is to find the current ( $I$ ), so let's isolate it!

STEP 8

Notice that the length of the wire ( $L$ ) appears on both sides of the equation.
We can divide both sides by $L$ to get rid of it: $\frac{I \cdot L \cdot B}{L} = \frac{\mu \cdot L \cdot g}{L}$ .
This simplifies to $I \cdot B = \mu \cdot g$ .

STEP 9

Now, we can divide both sides by the magnetic field strength ( $B$ ) to solve for the current: $\frac{I \cdot B}{B} = \frac{\mu \cdot g}{B}$ , which gives us $I = \frac{\mu \cdot g}{B}$ .

STEP 10

Let's plug in the given values: $\mu = 0.010 \, \text{kg/m}$ , $g = 9.8 \, \text{m/s}^2$ , and $B = 0.000020 \, \text{T}$ .

STEP 11

So, $I = \frac{0.010 \, \text{kg/m} \cdot 9.8 \, \text{m/s}^2}{0.000020 \, \text{T}} = \frac{0.098 \, \text{kg} \cdot \text{m/s}^2 \cdot \frac{1}{\text{m}}}{0.000020 \, \text{T}} = \frac{0.098 \, \text{N}}{0.000020 \, \text{T}} = 4900 \, \text{A}$ .
Remember, $1 \, \text{Tesla} = 1 \, \frac{\text{N}}{\text{A} \cdot \text{m}}$ , and since the length canceled out earlier, the units work out perfectly to amperes!

STEP 12

The required current is 4900 A.
With this current, the wire would likely melt or sag significantly due to the heat generated by such a high current.

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Studdy Solution

STEP 1

STEP 2

1. Formula for Magnetic Force on a Wire
2. Calculate the Current

STEP 3

STEP 4

We want the magnetic force ( $F$ ) to be equal to the wire's weight, which is just mass ( $m$ ) times gravity ( $g$ ).
So, we have $F = m \cdot g$ .
Gravity is approximately $9.8 \, \text{m/s}^2$ .

STEP 5

We're given the linear density ( $\mu$ ), which is the mass per unit length: $\mu = \frac{m}{L}$ .
We can rewrite this as $m = \mu \cdot L$ .

STEP 6

Now, we can substitute this expression for mass into our weight equation: $F = (\mu \cdot L) \cdot g$ .
See how the length ( $L$ ) is now involved?
This is great because it also appears in the magnetic force formula!

STEP 7

Now, we can set the magnetic force equal to the weight: $I \cdot L \cdot B = \mu \cdot L \cdot g$ .
Our goal is to find the current ( $I$ ), so let's isolate it!

STEP 8

Notice that the length of the wire ( $L$ ) appears on both sides of the equation.
We can divide both sides by $L$ to get rid of it: $\frac{I \cdot L \cdot B}{L} = \frac{\mu \cdot L \cdot g}{L}$ .
This simplifies to $I \cdot B = \mu \cdot g$ .

STEP 9

Now, we can divide both sides by the magnetic field strength ( $B$ ) to solve for the current: $\frac{I \cdot B}{B} = \frac{\mu \cdot g}{B}$ , which gives us $I = \frac{\mu \cdot g}{B}$ .

STEP 10

Let's plug in the given values: $\mu = 0.010 \, \text{kg/m}$ , $g = 9.8 \, \text{m/s}^2$ , and $B = 0.000020 \, \text{T}$ .

STEP 11

STEP 12

The required current is 4900 A.
With this current, the wire would likely melt or sag significantly due to the heat generated by such a high current.

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Studdy Solution

STEP 1

STEP 2

1. Formula for Magnetic Force on a Wire2. Calculate the Current

STEP 3

STEP 4

We want the magnetic force (FFF) to be equal to the wire's weight, which is just mass (mmm) times gravity (ggg).So, we have F=m⋅gF = m \cdot gF=m⋅g.Gravity is approximately 9.8 m/s29.8 \, \text{m/s}^29.8m/s2.

STEP 5

We're given the *linear density* (μ\muμ), which is the mass per unit length: μ=mL\mu = \frac{m}{L}μ=Lm​.We can rewrite this as m=μ⋅Lm = \mu \cdot Lm=μ⋅L.

STEP 6

Now, we can substitute this expression for mass into our weight equation: F=(μ⋅L)⋅gF = (\mu \cdot L) \cdot gF=(μ⋅L)⋅g.See how the length (LLL) is now involved?This is great because it also appears in the magnetic force formula!

STEP 7

Now, we can set the magnetic force equal to the weight: I⋅L⋅B=μ⋅L⋅gI \cdot L \cdot B = \mu \cdot L \cdot gI⋅L⋅B=μ⋅L⋅g.Our goal is to find the current (III), so let's **isolate** it!

STEP 8

STEP 9

Now, we can divide both sides by the magnetic field strength (BBB) to **solve for** the current: I⋅BB=μ⋅gB\frac{I \cdot B}{B} = \frac{\mu \cdot g}{B}BI⋅B​=Bμ⋅g​, which gives us I=μ⋅gBI = \frac{\mu \cdot g}{B}I=Bμ⋅g​.

STEP 10

Let's **plug in** the given values: μ=0.010 kg/m\mu = 0.010 \, \text{kg/m}μ=0.010kg/m, g=9.8 m/s2g = 9.8 \, \text{m/s}^2g=9.8m/s2, and B=0.000020 TB = 0.000020 \, \text{T}B=0.000020T.

STEP 11

STEP 12

The required current is **4900 A**.With this current, the wire would likely **melt** or **sag** significantly due to the heat generated by such a high current.

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Studdy solves anything!

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Formula for Magnetic Force on a Wire
2. Calculate the Current

We want the magnetic force ( $F$ ) to be equal to the wire's weight, which is just mass ( $m$ ) times gravity ( $g$ ).
So, we have $F = m \cdot g$ .
Gravity is approximately $9.8 \, \text{m/s}^2$ .

We're given the linear density ( $\mu$ ), which is the mass per unit length: $\mu = \frac{m}{L}$ .
We can rewrite this as $m = \mu \cdot L$ .

Now, we can substitute this expression for mass into our weight equation: $F = (\mu \cdot L) \cdot g$ .
See how the length ( $L$ ) is now involved?
This is great because it also appears in the magnetic force formula!

Now, we can set the magnetic force equal to the weight: $I \cdot L \cdot B = \mu \cdot L \cdot g$ .
Our goal is to find the current ( $I$ ), so let's isolate it!

Now, we can divide both sides by the magnetic field strength ( $B$ ) to solve for the current: $\frac{I \cdot B}{B} = \frac{\mu \cdot g}{B}$ , which gives us $I = \frac{\mu \cdot g}{B}$ .

Let's plug in the given values: $\mu = 0.010 \, \text{kg/m}$ , $g = 9.8 \, \text{m/s}^2$ , and $B = 0.000020 \, \text{T}$ .

The required current is 4900 A.
With this current, the wire would likely melt or sag significantly due to the heat generated by such a high current.