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Physics II - B15

TABLE OF CONTENTS

Review Ampere’s Law

  • Infinite current sheet:
\[B = \frac{\mu_0 J}2\]
  • Two parallel infinite current sheets:
\[B = \mu_0J\]
  • Solenoid:
\[B = \mu_0 n J\]

Magnetic Flux

Gauss’ Law in Magnetism

Assignments - Chapter 29

\[B_B > B_C > B_A = 0\]
\[\displaystyle B = \frac{\mu_0I}{2r} = \frac{\mu_0\frac{e}{T}}{2r} = \frac{\mu_0\frac{ev}{2\pi r}}{2r} = 12.5(T)\]
\[\vec B_{total} = \vec B_{straight} + \vec B_{circular}\] \[B_{total} = B_{straight} + B_{circular}\] \[B_{total}= \frac{\mu_0I}{2r} + \frac{\mu_0I}{2\pi r} = 5.52 \times 10^{-6} (T)\]


  • The direction of forces can be deduced from the direction of $I_1$ and the magnetic field, as in image.
  • We have:
\[\vec F_{net} = \vec F_{AB} + \vec F_{BC} + \vec F_{CD} + \vec F_{DA}\]
  • $F_{BC}$ and $F_{AD}$ are of the same magnitude and opposite directions. We have:
\[\vec F_{BC} + \vec F_{AD} = 0\]
  • Therefore:
\[F_{net} = F_{AB} - F_{CD} = \frac{\mu_0I_1I_2l}{2\pi}\bigg(\frac{1}{c} - \frac{1}{c+a}\bigg) = 2.7 \times 10^{-5} (N)\]
\[c > a > d > b\]


$a)$ The magnitude and direction of the magnetic field at point $a$ is:

\[B_a = \frac{\mu_0I_1}{2\pi d} = 2 \times 10^{-4} (T)\]

The direction of $B_a$ would be upward.

$b)$ The magnitude and direction of the magnetic field at point $b$ is:

\[B_b = \frac{\mu_0(I_1-I_2)}{2\pi d} = -1.3 \times 10^{-4} (T)\]

The direction of $B_b$ would be downward.