Rowan-Classes/5th-Semester-Fall-2023/EEMAGS/Homework/Homework-02.md

Homework 2 - Aidan Sharpe

1

Consider a long cylindrical wire of radius, a, carrying a current I = I_0 \cos(\omega t)\hat{z}.

a)

Write an expression for the magnetic field strength, B, outside the wire (\rho > a):

By Ampere's Law:

\int \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}

For a closed loop of radius, \rho:

\int \vec{B} \cdot d\vec{l} = B(2\pi\rho)

\therefore B(2\pi\rho) = \mu_0 I_{enc}

Since (\rho > a):

I_{enc} = I = I_0 \cos(\omega t)

Recombining:

B(2\pi\rho) = \mu_0 I_0 \cos(\omega t)

\therefore B = {\mu_0 I_0 \cos(\omega t) \over 2\pi\rho} \text{[T]}

b)

Consider a rectangular loop a distance, d, from the wire with sidelengths \alpha in the \hat{x} direction, and \beta in the \hat{z} direction.

i)

Calculate the magnetic flux, \Phi_B, through the loop.

By Gauss's Law for magnetism:

\Phi_B = \iint \vec{B} \cdot d\vec{s}

Since the \beta does not vary:

\Phi_B = \beta \int\limits_{d}^{d+\alpha} {\mu_0 I_0 \cos(\omega t) \over 2\pi\rho} d\rho

Taking out constants:

\Phi_B = {\beta \mu_0 I_0 \cos(\omega t) \over 2 \pi} \int\limits_d^{d+\alpha} {1 \over \rho}d\rho

Evaluate:

\Phi_B = {\beta \mu_0 I_0 \cos(\omega t) \over 2\pi} \ln\left( {\lvert d + \alpha \rvert \over \lvert d \rvert} \right) \text{[Vs]}

ii)

Find the induced EMF, \cal{E}:

\mathcal{E} = -{d \over dt}\Phi_B

Plug in:

\mathcal{E} = -{d \over dt} {\beta \mu_0 I_0 \cos(\omega t) \over 2\pi} \ln\left({\lvert d + \alpha \rvert \over \lvert d \rvert}\right)

Evaluate:

\mathcal{E} = {\beta \mu_0 I_0 \sin(\omega t) \over 2\pi} \ln\left({\lvert d + \alpha \rvert \over \lvert d \rvert}\right) \text{[V]}

2

A quarter circle loop of wire with inner radius, a, and outer radius, b, has current, I. Find the magnetic field strength at the center of the circle.

By superposition:

B = B_1 + B_2 + B_3 + B_4

By Ampere's law:

\int \vec{B_1} \cdot d\vec{l} = \mu_0 I_{enc_1}

\int \vec{B_2} \cdot d\vec{l} = \mu_0 I_{enc_2}

\int \vec{B_3} \cdot d\vec{l} = \mu_0 I_{enc_3}

\int \vec{B_4} \cdot d\vec{l} = \mu_0 I_{enc_4}

Since the current in the first and third segments are either parallel or antiparallel:

I_{enc_1} = 0

I_{enc_3} = 0

Since all of I is enclosed in a loop from either segment two or four:

I_{enc_2} = I_{enc_4} = I

Back to Ampere's Law:

\int \vec{B_1} \cdot d\vec{l} = 0 \therefore B_1 = 0

\int \vec{B_2} \cdot d\vec{l} = \mu_0 I

\int \vec{B_3} \cdot d\vec{l} = 0 \therefore B_2 = 0

\int \vec{B_4} \cdot d\vec{l} = \mu_0 I

Determining d\vec{l} for segments two and four:

\int \vec{B_2} \cdot \vec{a}d\varphi = \mu_0 I

\int \vec{B_4} \cdot \vec{b}d\varphi = \mu_0 I

Determining the bounds:

\int\limits_0^{\pi \over 2} \vec{B_2} \cdot \vec{a} d\varphi = \mu_0 I

\int\limits^0_{\pi \over 2} \vec{B_4} \cdot \vec{b} d\varphi = \mu_0 I

Evaluate:

B_2 \left({\pi a \over 2}\right) = \mu_0 I

B_4 \left({-\pi b \over 2}\right) = \mu_0 I

Solve for B:

B_2 = {2 \mu_0 I \over \pi a}

B_4 = -{2 \mu_0 I \over \pi b}

B = {2 \mu_0 I \over \pi a} - {2 \mu_0 I \over \pi b} = {2 \mu_0 I (a-b) \over \pi^2 a b} \text{[T]}

3

Consider a cylinder of radius, \rho_0 = 0.5\text{[m]}, with a current density, \vec{J} = 4.5e^{-2\rho}\hat{z}[A/m^2].

By Ampere's Law:

\int \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}

For a radial current density:

I_{enc} = \int\limits_{0}^{2\pi} \int\limits_{0}^{\rho} J sdsd\varphi

Plugging in J with s as the integrating variable:

I_{enc} = \int\limits_{0}^{2\pi} \int\limits_{0}^{\rho} 4.5e^{-2s} sdsd\varphi

I_{enc} = 4.5\int\limits_{0}^{2\pi} \int\limits_{0}^{\rho} e^{-2s} sdsd\varphi

I_{enc} = 4.5\int\limits_{0}^{2\pi} \left[ \left( {-\rho \over 2} - {1 \over 4}\right)e^{-2\rho} + {1 \over 4}\right]d\varphi

\therefore I_{enc} = {9 \pi e^{-2\rho}(e^{2\rho} -2\rho -1) \over 4}

Back to Ampere's Law:

\int \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}

For a circular loop:

\int\limits_{0}^{2\pi} B\rho d\varphi = \mu_0 I_{enc}

B(2\pi\rho) = {9 \pi e^{-2\rho}(e^{2\rho} -2\rho -1) \over 4}

For \rho \le \rho_0

B = {9 e^{-2\rho}(e^{2\rho} -2\rho -1) \over 8\rho}

For \rho \ge \rho_0:

I_{enc} = 4.5\int\limits_{0}^{2\pi} \int\limits_{0}^{\rho_0} e^{-2s} sdsd\varphi = {9 (e-2) \pi \over 4 e}

B(2\pi\rho) = {9(e-2)\pi \over 4e}

\therefore B = {9(e-2) \over 8e\rho}\text{[T]}

4

Consider a solenoidal wire with n coils per unit length. The core becomes magnetized when a current I=10[A] is put into the wire coil, and this causes a bound current to flow around the cylindrical surface of the core as shown in the side view diagram. This bound core surface current density has magnitude, J = 20n [A/m]

B = B_s + B_c

B_s = \mu_0 I n = 10\mu_0n

Pretend the core current is just another solenoid:

B_c = \mu_0 I n = 20\mu_0n

B = 30\mu_0 n \text{[T]}

5

A satellite travelling at 5\text{[km/s]} enters a current filled curtain. From t=1\text{[s]} to t=3\text{[s]}, the satellite's magnetometer increases from -95\hat{x}\text{[nT]} to 95\hat{x}\text{[nT]}. If the current flows in the \hat{z} direction, find the current density, J.

By Ampere's Law:

\int \vec{B} \cdot d\vec{l} = \mu_0 \iint \vec{J} \cdot d\vec{s}

The total increase in B is 195 \text{[nT]}, so:

195 \times 10^{-9} = \mu_0 \iint \vec{J} \cdot d\vec{s}

Plug in bounds to get the total distance through the curtain:

195 \times 10^{-9} = \mu_0 \int\limits_{1}^{3} \int\limits_{0}^{-5000} J dvdt

195 \times 10^{-9} = -10000J\mu_0

J = {-195 \times 10^{-13} \over \mu_0} \text{[A/m}]