Magnetism flash cards

1. 1.What did Oersted's experiment (1820) demonstrate?

   A current-carrying wire deflects a nearby magnetic compass needle, proving that an electric current produces a magnetic field around it. The deflection reverses when current direction reverses.

   Hint: Compass near a wire.

2. 2.State the Biot–Savart law for the field of a current element.

   $d\vec{B}=\dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{l}\times\hat{r}}{r^2}$. Magnitude $dB=\dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$, where $\theta$ is the angle between $d\vec{l}$ and $\hat r$.

   Hint: Inverse-square, cross product.

3. 3.What is the value of $\mu_0$ and its SI unit?

   $\mu_0=4\pi\times10^{-7}\ \text{T m A}^{-1}$ (also written $\text{H m}^{-1}$). It is the permeability of free space.

   Hint: $4\pi\times10^{-7}$.

4. 4.For a current element, where is $dB$ maximum and where zero?

   $dB\propto\sin\theta$. It is maximum at $\theta=90^\circ$ (point perpendicular to the element) and zero along the axis of the element ($\theta=0^\circ$ or $180^\circ$).

   Hint: $\sin\theta$ factor.

5. 5.Magnetic field at distance $a$ from a long straight current-carrying wire?

   $B=\dfrac{\mu_0 I}{2\pi a}$, circular field lines around the wire; direction from right-hand rule.

   Hint: $1/a$ dependence.

6. 6.Field due to a straight wire of finite length (angles $\theta_1,\theta_2$ at the ends from the foot of perpendicular)?

   $B=\dfrac{\mu_0 I}{4\pi a}(\sin\theta_1+\sin\theta_2)$, where $a$ is the perpendicular distance and angles are measured from the perpendicular to the lines joining point to the ends.

   Hint: Reduces to $\mu_0 I/2\pi a$ for infinite wire.

7. 7.State the right-hand thumb rule for a straight wire.

   Grip the wire with the right hand, thumb pointing along the current; the curl of the fingers gives the direction of the circular magnetic field lines.

   Hint: Thumb = current.

8. 8.Magnetic field at the centre of a circular current loop of radius $R$?

   $B=\dfrac{\mu_0 I}{2R}$, directed along the axis (right-hand rule). For $N$ turns, $B=\dfrac{\mu_0 N I}{2R}$.

   Hint: Half of $\mu_0 I/R$.

9. 9.Field on the axis of a circular loop at distance $x$ from centre?

   $B=\dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ (for $N$ turns multiply by $N$). Directed along the axis.

   Hint: Peaks at $x=0$.

10. 10.Far on the axis ($x\gg R$), how does a loop's field behave?

    $B\approx\dfrac{\mu_0}{2\pi}\dfrac{IA}{x^3}=\dfrac{\mu_0}{2\pi}\dfrac{m}{x^3}$ with $m=IA$ — identical to a magnetic dipole field on its axis.

    Hint: Loop = dipole, $1/x^3$.

11. 11.Field at the centre of a circular arc subtending angle $\phi$ (radians) at radius $R$?

    $B=\dfrac{\mu_0 I}{4\pi R}\,\phi=\dfrac{\mu_0 I \phi}{4\pi R}$. For a full circle $\phi=2\pi$ gives $\mu_0 I/2R$.

    Hint: Fraction $\phi/2\pi$ of a full loop.

12. 12.Field at centre of a semicircular arc of radius $R$?

    $B=\dfrac{\mu_0 I}{4R}$ (half the full-loop value), perpendicular to the plane of the arc.

    Hint: $\phi=\pi$.

13. 13.State Ampere's circuital law.

    $\oint\vec{B}\cdot d\vec{l}=\mu_0 I_{enc}$: the line integral of $\vec B$ around any closed loop equals $\mu_0$ times the net current threading (enclosed by) that loop.

    Hint: Magnetic analogue of Gauss's law.

14. 14.Use Ampere's law to get the field of a long straight wire.

    Take a circular Amperian loop of radius $a$: $B(2\pi a)=\mu_0 I\Rightarrow B=\dfrac{\mu_0 I}{2\pi a}$. Symmetry makes $\vec B$ tangential and uniform on the loop.

    Hint: Circle of radius $a$, $\oint dl=2\pi a$.

15. 15.Field inside and outside a long solenoid ($n$ = turns per unit length)?

    Inside (near the middle): $B=\mu_0 n I$, uniform and axial. Outside an ideal long solenoid: $B\approx0$.

    Hint: Depends on $n$, not radius, inside.

16. 16.Field at the end (mouth) of a long solenoid?

    $B=\dfrac{1}{2}\mu_0 n I$ — half the value at the middle, because the end sees roughly half the winding.

    Hint: Half of the interior value.

17. 17.Field inside a toroid (mean radius $r$, $N$ total turns)?

    $B=\dfrac{\mu_0 N I}{2\pi r}=\mu_0 n I$ with $n=N/2\pi r$. Field is confined within the core; outside (and in the central hole) $B=0$.

    Hint: Solenoid bent into a doughnut.

18. 18.Why is $B=0$ outside an ideal toroid?

    An Amperian loop outside the core encloses zero net current (equal currents up and down through the plane cancel), so $\oint\vec B\cdot d\vec l=0$.

    Hint: Enclosed current cancels.

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