QuestionWhat is the current in a wire of radius $R=3.61 \mathrm{~mm}$ if the magnitude of the current density is given by $(a) J_{a}=J_{0} r / R$ and $(b) J_{b}=J_{0}(1-r / R)$ in which $r$ is the radial distance and $J_{0}=4.03 \times 10^{4} \mathrm{~A} / \mathrm{m}^{2}$ ? (c) Which function maximizes the current density near the wire's surface? (a) Number i Units $\square$ (b) Number i $\square$ Units $\square$

Studdy Solution

STEP 1

What is this asking? We need to figure out the total current flowing through a wire given two different ways the current density changes as you move from the center of the wire outwards.
Then, we need to determine which current density function leads to a higher current density near the wire's surface. Watch out! Current and current density aren't the same thing!
Current density is current per unit area.
Don't mix up radius and area either.
Also, don't forget units!

STEP 2

1. Calculate Current with the First Density Function
2. Calculate Current with the Second Density Function
3. Maximize Current Density

STEP 3

Alright, let's define our first current density function: $J_a = J_0 \frac{r}{R}$ .
Remember, this tells us how much current is squeezed through a tiny bit of area at a distance $r$ from the center.

STEP 4

To get the total current, we need to integrate this density over the entire cross-sectional area of the wire.
Think of slicing the wire into thin rings of radius $r$ and thickness $dr$ .
The area of such a ring is $dA = 2\pi r dr$ .

STEP 5

Now, the current through each ring is $dI_a = J_a dA$ .
Let's substitute our expressions: $dI_a = \left(J_0 \frac{r}{R}\right)(2\pi r dr) = \frac{2\pi J_0}{R} r^2 dr$ .

STEP 6

Time to integrate!
We'll integrate from $r = 0$ (the center) to $r = R$ (the surface): $I_a = \int_0^R dI_a = \int_0^R \frac{2\pi J_0}{R} r^2 dr = \frac{2\pi J_0}{R} \int_0^R r^2 dr = \frac{2\pi J_0}{R} \left[ \frac{r^3}{3} \right]_0^R = \frac{2\pi J_0}{R} \cdot \frac{R^3}{3} = \frac{2\pi J_0 R^2}{3}.$

STEP 7

Let's plug in the numbers: $R = \textbf{3.61 mm} = \textbf{0.00361 m}$ and $J_0 = \textbf{4.03} \cdot \textbf{10}^4 \textbf{A/m}^2$ . $I_a = \frac{2\pi (\textbf{4.03} \cdot \textbf{10}^4 \textbf{A/m}^2) (\textbf{0.00361 m})^2}{3} \approx \textbf{1.10 A}.$

STEP 8

Our second current density function is $J_b = J_0\left(1 - \frac{r}{R}\right)$ .
We'll follow the same process as before.

STEP 9

The current through each thin ring is $dI_b = J_b dA = J_0\left(1 - \frac{r}{R}\right) 2\pi r dr = 2\pi J_0 \left(r - \frac{r^2}{R}\right) dr$ .

STEP 10

Integrating from $r = 0$ to $r = R$ : $I_b = \int_0^R 2\pi J_0 \left(r - \frac{r^2}{R}\right) dr = 2\pi J_0 \left[ \frac{r^2}{2} - \frac{r^3}{3R} \right]_0^R = 2\pi J_0 \left(\frac{R^2}{2} - \frac{R^3}{3R}\right) = 2\pi J_0 R^2 \left(\frac{1}{2} - \frac{1}{3}\right) = \frac{\pi J_0 R^2}{3}.$

STEP 11

Substituting the values: $I_b = \frac{\pi (\textbf{4.03} \cdot \textbf{10}^4 \textbf{A/m}^2) (\textbf{0.00361 m})^2}{3} \approx \textbf{0.550 A}.$

STEP 12

We want to know which function, $J_a$ or $J_b$ , gives a larger current density near the wire's surface. "Near the surface" means $r$ is close to $R$ .

STEP 13

Let's evaluate both functions at $r = R$ : $J_a(R) = J_0 \frac{R}{R} = J_0$ and $J_b(R) = J_0\left(1 - \frac{R}{R}\right) = 0$ .

STEP 14

Clearly, $J_a$ gives a larger current density near the surface.

STEP 15

(a) The current for the first density function is $\textbf{1.10 A}$ . (b) The current for the second density function is $\textbf{0.550 A}$ . (c) The function that maximizes the current density near the wire's surface is $J_a = J_0 \frac{r}{R}$ .

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Studdy Solution

STEP 1

STEP 2

1. Calculate Current with the First Density Function
2. Calculate Current with the Second Density Function
3. Maximize Current Density

STEP 3

Alright, let's define our first current density function: $J_a = J_0 \frac{r}{R}$ .
Remember, this tells us how much current is squeezed through a tiny bit of area at a distance $r$ from the center.

STEP 4

To get the total current, we need to integrate this density over the entire cross-sectional area of the wire.
Think of slicing the wire into thin rings of radius $r$ and thickness $dr$ .
The area of such a ring is $dA = 2\pi r dr$ .

STEP 5

Now, the current through each ring is $dI_a = J_a dA$ .
Let's substitute our expressions: $dI_a = \left(J_0 \frac{r}{R}\right)(2\pi r dr) = \frac{2\pi J_0}{R} r^2 dr$ .

STEP 6

STEP 7

Let's plug in the numbers: $R = \textbf{3.61 mm} = \textbf{0.00361 m}$ and $J_0 = \textbf{4.03} \cdot \textbf{10}^4 \textbf{A/m}^2$ . $I_a = \frac{2\pi (\textbf{4.03} \cdot \textbf{10}^4 \textbf{A/m}^2) (\textbf{0.00361 m})^2}{3} \approx \textbf{1.10 A}.$

STEP 8

Our second current density function is $J_b = J_0\left(1 - \frac{r}{R}\right)$ .
We'll follow the same process as before.

STEP 9

The current through each thin ring is $dI_b = J_b dA = J_0\left(1 - \frac{r}{R}\right) 2\pi r dr = 2\pi J_0 \left(r - \frac{r^2}{R}\right) dr$ .

STEP 10

Integrating from $r = 0$ to $r = R$ : $I_b = \int_0^R 2\pi J_0 \left(r - \frac{r^2}{R}\right) dr = 2\pi J_0 \left[ \frac{r^2}{2} - \frac{r^3}{3R} \right]_0^R = 2\pi J_0 \left(\frac{R^2}{2} - \frac{R^3}{3R}\right) = 2\pi J_0 R^2 \left(\frac{1}{2} - \frac{1}{3}\right) = \frac{\pi J_0 R^2}{3}.$

STEP 11

Substituting the values: $I_b = \frac{\pi (\textbf{4.03} \cdot \textbf{10}^4 \textbf{A/m}^2) (\textbf{0.00361 m})^2}{3} \approx \textbf{0.550 A}.$

STEP 12

We want to know which function, $J_a$ or $J_b$ , gives a larger current density near the wire's surface. "Near the surface" means $r$ is close to $R$ .

STEP 13

Let's evaluate both functions at $r = R$ : $J_a(R) = J_0 \frac{R}{R} = J_0$ and $J_b(R) = J_0\left(1 - \frac{R}{R}\right) = 0$ .

STEP 14

Clearly, $J_a$ gives a larger current density near the surface.

STEP 15

(a) The current for the first density function is $\textbf{1.10 A}$ . (b) The current for the second density function is $\textbf{0.550 A}$ . (c) The function that maximizes the current density near the wire's surface is $J_a = J_0 \frac{r}{R}$ .

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Studdy Solution

STEP 1

STEP 2

1. Calculate Current with the First Density Function2. Calculate Current with the Second Density Function3. Maximize Current Density

STEP 3

Alright, let's **define** our first current density function: Ja=J0rRJ_a = J_0 \frac{r}{R}Ja​=J0​Rr​.Remember, this tells us how much current is squeezed through a tiny bit of area at a distance rrr from the center.

STEP 4

To get the total current, we need to **integrate** this density over the entire cross-sectional area of the wire.Think of slicing the wire into thin rings of radius rrr and thickness drdrdr.The area of such a ring is dA=2πrdrdA = 2\pi r drdA=2πrdr.

STEP 5

Now, the current through each ring is dIa=JadAdI_a = J_a dAdIa​=Ja​dA.Let's **substitute** our expressions: dIa=(J0rR)(2πrdr)=2πJ0Rr2drdI_a = \left(J_0 \frac{r}{R}\right)(2\pi r dr) = \frac{2\pi J_0}{R} r^2 drdIa​=(J0​Rr​)(2πrdr)=R2πJ0​​r2dr.

STEP 6

STEP 7

STEP 8

Our second current density function is Jb=J0(1−rR)J_b = J_0\left(1 - \frac{r}{R}\right)Jb​=J0​(1−Rr​).We'll follow the same process as before.

STEP 9

The current through each thin ring is dIb=JbdA=J0(1−rR)2πrdr=2πJ0(r−r2R)drdI_b = J_b dA = J_0\left(1 - \frac{r}{R}\right) 2\pi r dr = 2\pi J_0 \left(r - \frac{r^2}{R}\right) drdIb​=Jb​dA=J0​(1−Rr​)2πrdr=2πJ0​(r−Rr2​)dr.

STEP 10

STEP 11

**Substituting** the values: Ib=π(4.03⋅104A/m2)(0.00361 m)23≈0.550 A.I_b = \frac{\pi (\textbf{4.03} \cdot \textbf{10}^4 \textbf{A/m}^2) (\textbf{0.00361 m})^2}{3} \approx \textbf{0.550 A}.Ib​=3π(4.03⋅104A/m2)(0.00361 m)2​≈0.550 A.

STEP 12

We want to know which function, JaJ_aJa​ or JbJ_bJb​, gives a larger current density near the wire's surface. "Near the surface" means rrr is close to RRR.

STEP 13

Let's **evaluate** both functions at r=Rr = Rr=R: Ja(R)=J0RR=J0J_a(R) = J_0 \frac{R}{R} = J_0Ja​(R)=J0​RR​=J0​ and Jb(R)=J0(1−RR)=0J_b(R) = J_0\left(1 - \frac{R}{R}\right) = 0Jb​(R)=J0​(1−RR​)=0.

STEP 14

Clearly, JaJ_aJa​ gives a larger current density near the surface.

STEP 15

(a) The current for the first density function is 1.10 A\textbf{1.10 A}1.10 A. (b) The current for the second density function is 0.550 A\textbf{0.550 A}0.550 A. (c) The function that maximizes the current density near the wire's surface is Ja=J0rRJ_a = J_0 \frac{r}{R}Ja​=J0​Rr​.

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1. Calculate Current with the First Density Function
2. Calculate Current with the Second Density Function
3. Maximize Current Density

Alright, let's define our first current density function: $J_a = J_0 \frac{r}{R}$ .
Remember, this tells us how much current is squeezed through a tiny bit of area at a distance $r$ from the center.

To get the total current, we need to integrate this density over the entire cross-sectional area of the wire.
Think of slicing the wire into thin rings of radius $r$ and thickness $dr$ .
The area of such a ring is $dA = 2\pi r dr$ .

Now, the current through each ring is $dI_a = J_a dA$ .
Let's substitute our expressions: $dI_a = \left(J_0 \frac{r}{R}\right)(2\pi r dr) = \frac{2\pi J_0}{R} r^2 dr$ .

Our second current density function is $J_b = J_0\left(1 - \frac{r}{R}\right)$ .
We'll follow the same process as before.

The current through each thin ring is $dI_b = J_b dA = J_0\left(1 - \frac{r}{R}\right) 2\pi r dr = 2\pi J_0 \left(r - \frac{r^2}{R}\right) dr$ .

Substituting the values: $I_b = \frac{\pi (\textbf{4.03} \cdot \textbf{10}^4 \textbf{A/m}^2) (\textbf{0.00361 m})^2}{3} \approx \textbf{0.550 A}.$

We want to know which function, $J_a$ or $J_b$ , gives a larger current density near the wire's surface. "Near the surface" means $r$ is close to $R$ .

Let's evaluate both functions at $r = R$ : $J_a(R) = J_0 \frac{R}{R} = J_0$ and $J_b(R) = J_0\left(1 - \frac{R}{R}\right) = 0$ .

Clearly, $J_a$ gives a larger current density near the surface.

(a) The current for the first density function is $\textbf{1.10 A}$ . (b) The current for the second density function is $\textbf{0.550 A}$ . (c) The function that maximizes the current density near the wire's surface is $J_a = J_0 \frac{r}{R}$ .