In: Physics

# The figure below shows a cross section across a diameter of a long cylindrical conductor

The figure below shows a cross section across a diameter of a long cylindrical conductor of radius a = 2.00 cm carrying uniform current 159 A. What is the magnitude of the current's magnetic field at the following radial distances?

(a) 0
_____ T

(b) 1.20 cm
_____ T

(c) 2.00 cm (wire's surface)
______ T

(d) 3.90 cm
______ T

## Solutions

##### Expert Solution

Concepts and reason

The required concepts to solve these questions are ampere's law and magnetic field due to long cylindrical conductor with uniform current. Initially, use the expression for the magnetic field inside the long cylindrical conductor to calculate magnetic field for the given condition. Next, use the expression for the magnetic field inside the long cylindrical conductor to calculate the magnetic field inside the long cylindrical conductor with uniform current when $$r=1.20 \mathrm{~cm} .$$ Further, use expression for the magnetic field inside the long cylindrical conductor to calculate the magnetic field for the given condition. And, finally, use the expression for magnetic field outside the cylindrical conductor to calculate the magnetic field for the given condition.

Fundamentals

The expression for magnetic field inside the long cylindrical conductor with uniform current is, $$B_{\mathrm{IN}}=\frac{\mu_{0} I r}{2 \pi a^{2}}$$

Here, $$I$$ is current, $$a$$ is the radius of long cylinder conductor, $$r$$ is the radius inside the cylinder and $$\mu_{0}$$ is the is the permeability. The expression for magnetic field outside the cylindrical conductor is, $$B_{\mathrm{OUT}}=\frac{\mu_{0} I}{2 \pi r}$$

Here, $$I$$ is current, $$r$$ is the radius inside the cylinder and $$\mu_{0}$$ is the permeability.

(a)

Write the expression for magnetic field inside the long cylindrical conductor. $$B_{\mathrm{IN}}=\frac{\mu_{0} I r}{2 \pi a^{2}}$$

Consider, the value of the permeability $$\mu_{0}$$ is $$\left(1.257 \times 10^{-6} \mathrm{H} / \mathrm{m}\right)$$ Substitute $$\left(1.257 \times 10^{-6} \mathrm{H} / \mathrm{m}\right)$$ for $$\mu_{0},(2 \mathrm{~cm})$$ for $$a, 159 \mathrm{~A}$$ for $$I$$ and 0 for $$r$$

$$\begin{array}{c} B_{\mathrm{IN}}=\frac{\left(1.257 \times 10^{-6} \mathrm{H} / \mathrm{m}\right)(159 \mathrm{~A})(0)}{2 \pi(2 \mathrm{~cm})^{2}} \\ =0 \mathrm{~T} \end{array}$$

Part a

The magnitude of magnetic field when $$r=0$$ is $$0 \mathrm{~T}$$.

Since the given radial distance is zero, such that $$r=0$$, it is clear that the magnetic field is at the center of the cylindrical conductor which has no magnetic effect. The magnetic field for the given condition is to be calculated. Use the expression for the magnetic field inside the cylindrical conductor. Substitute the required values in the expression and solve for the magnetic field for the given radial distance.

(b)

Write the expression for magnetic field inside the long cylindrical conductor. $$B_{\mathrm{IN}}=\frac{\mu 0 I r}{2 \pi a^{2}}$$

Substitute $$\left(1.257 \times 10^{-6} \mathrm{H} / \mathrm{m}\right)$$ for $$\mu_{0},(2 \mathrm{~cm})$$ for $$a, 159$$ Afor $$I$$ and $$(1.20 \mathrm{~cm})$$ for $$r$$

$$\begin{array}{c} B_{\mathrm{IN}}=\frac{\left(1.257 \times 10^{-6} \mathrm{H} / \mathrm{m}\right)(159 \mathrm{~A})\left(1.20 \mathrm{~cm}\left(\frac{1 \mathrm{~m}}{10^{2} \mathrm{~cm}}\right)\right)}{2 \pi\left(2 \mathrm{~cm}\left(\frac{1 \mathrm{~m}}{10^{2} \mathrm{~cm}}\right)\right)^{2}} \\ =\frac{2.398 \times 10^{-6} \mathrm{~A} \cdot \mathrm{H}}{0.251 \times 10^{-2} \mathrm{~m}^{2}} \\ =9.54 \times 10^{-4} \mathrm{~T} \end{array}$$

Part b The magnitude of magnetic field when $$r=1.20 \mathrm{cmis} 9.54 \times 10^{-4} \mathrm{~T}$$.

Since the given radial distance is less than the radius of the cylindrical conductor, such that $$r<a$$, it is clear that the magnetic field is inside the cylindrical conductor. The magnetic field inside the long cylindrical conductor is to be calculated. Use the expression for the magnetic field inside the cylindrical conductor. Substitute the required values in the expression and solve for the magnetic field for the given radial distance.

(c)

Write the expression for magnetic field inside the long cylindrical conductor. $$B_{\mathrm{IN}}=\frac{\mu_{0} I r}{2 \pi a^{2}}$$

Substitute $$\left(1.257 \times 10^{-6} \mathrm{H} / \mathrm{m}\right)$$ for $$\mu_{0},(2 \mathrm{~cm})$$ for $$a, 159$$ Afor $$I$$ and $$(2.00 \mathrm{~cm})$$ for $$r$$

$$\begin{array}{c} B_{\mathrm{IN}}=\frac{\left(1.257 \times 10^{-6} \mathrm{H} / \mathrm{m}\right)(159 \mathrm{~A})\left(2.00 \mathrm{~cm}\left(\frac{1 \mathrm{~m}}{10^{2} \mathrm{~cm}}\right)\right)}{2 \pi\left(2 \mathrm{~cm}\left(\frac{1 \mathrm{~m}}{10^{2} \mathrm{~cm}}\right)\right)^{2}} \\ =\frac{3.997 \times 10^{-6} \mathrm{~A} \cdot \mathrm{H}}{0.251 \times 10^{-2} \mathrm{~m}^{2}} \\ =1.59 \times 10^{-3} \mathrm{~T} \end{array}$$

Part c The magnitude of magnetic field when $$r=2 \mathrm{cmis} 1.59 \times 10^{-3} \mathrm{~T}$$

Since the given radial distance equals to the radius of the cylindrical conductor, such that $$r=a$$, it is clear that the magnetic field is on the surface of the cylindrical conductor. The magnetic field on the surface of the cylindrical conductor is to be calculated. Use the expression for the magnetic field inside the cylindrical conductor. Substitute the required values in the expression and solve for the magnetic field for the given radial distance.

(d)

Since the given radial distance is greater than the radius of the cylindrical conductor; $$r>a$$, it is clear that the magnetic field is outside the cylindrical conductor. Write the expression for magnetic field outside the cylindrical conductor. $$B_{\mathrm{OUT}}=\frac{\mu_{0} I}{2 \pi r}$$

Substitute $$\left(1.257 \times 10^{-6} \mathrm{H} / \mathrm{m}\right)$$ for $$\mu_{0}, 159$$ Afor $$I$$ and $$(3.90 \mathrm{~cm})$$ for $$r$$

\begin{aligned} B_{\mathrm{OUT}} &=\frac{\left(1.257 \times 10^{-6} \mathrm{H} / \mathrm{m}\right)(159 \mathrm{~A})}{2 \pi\left(3.90 \mathrm{~cm}\left(\frac{1 \mathrm{~m}}{10^{2} \mathrm{~cm}}\right)\right)} \\ &=\frac{199.86 \times 10^{-6} \mathrm{~A} \cdot \mathrm{H} / \mathrm{m}}{0.245 \times 10^{-2} \mathrm{~m}} \\ &=8.15 \times 10^{-4} \mathrm{~T} \end{aligned}

Part d The magnitude of magnetic field when $$r=2 \mathrm{cmis} 8.15 \times 10^{-4} \mathrm{~T}$$.

Since the given radial distance is greater than the radius of the cylindrical conductor, such that $$r>a$$, it is clear that the magnetic field is outside the cylindrical conductor. The magnetic field outside the long cylindrical conductor is to be calculated. Use the expression for the magnetic field outside the cylindrical conductor. Substitute the required values in the expression and solve for the magnetic field for the given radial distance.

Part a The magnitude of magnetic field when $$r=0$$ is $$0 \mathrm{~T}$$

Part b The magnitude of magnetic field when $$r=1.20 \mathrm{cmis} 9.54 \times 10^{-4} \mathrm{~T}$$.

Part c The magnitude of magnetic field when $$r=2 \mathrm{cmis} 1.59 \times 10^{-3} \mathrm{~T}$$.

Part d The magnitude of magnetic field when $$r=2 \mathrm{cmis} 8.15 \times 10^{-4} \mathrm{~T}$$