Magnetic Field

(c) magnetic moment of the coil

The magnetic dipole moment of a current loop is $m = NIA$ (unit A m$^2$), and its direction is along the normal to the plane of the coil. The torque on the coil in a field $B$ is $\tau = NIAB\sin\theta = mB\sin\theta$.

(a) Vector form of the Lorentz force

$$\vec F = q\,(\vec v \times \vec B)$$

Its magnitude is $F = qvB\sin\theta$, and it acts perpendicular to both $\vec v$ and $\vec B$. If an electric field is also present, the total Lorentz force is $\vec F = q(\vec E + \vec v \times \vec B)$.

(b) Fleming's left-hand rule

Stretch the forefinger, the middle finger and the thumb of the left hand mutually perpendicular to each other. If the forefinger points along the magnetic field and the middle finger along the current, then the thumb gives the direction of the force (motion) on the conductor.

(c) P.d. needed to make the wire self-supporting

Given:

• length of the wire, $l = 5$ cm $= 0.05$ m
• mass per unit length, $\lambda = 1.2$ g m$^{-1}$ $= 1.2 \times 10^{-3}$ kg m$^{-1}$
• magnetic flux density, $B = 0.6$ T (horizontal, perpendicular to the wire)
• resistance per unit length, $3.8\ \Omega$ m$^{-1}$
• $g = 9.8$ m s$^{-2}$

The wire is just self-supporting when the upward magnetic force equals its weight:

$$BIl = mg = \lambda l g \quad\Rightarrow\quad I = \frac{\lambda g}{B} = \frac{1.2 \times 10^{-3} \times 9.8}{0.6} = 1.96 \times 10^{-2}\ \text{A}$$

Resistance of the wire:

$$R = 3.8 \times 0.05 = 0.19\ \Omega$$

Required p.d.:

$$V = IR = 1.96 \times 10^{-2} \times 0.19 = 3.72 \times 10^{-3}\ \text{V}$$

The p.d. required is about $3.7 \times 10^{-3}$ V (3.7 mV). The current must flow in the direction that, by Fleming's left-hand rule, makes the force on the wire point upward.

(a) Biot–Savart law

The magnetic field $dB$ at a point P due to a small element $dl$ of a conductor carrying current $I$ is:

• directly proportional to the current, $dB \propto I$
• directly proportional to the length of the element, $dB \propto dl$
• directly proportional to $\sin\theta$, where $\theta$ is the angle between $d\vec l$ and the line joining the element to P
• inversely proportional to the square of the distance $r$ of P from the element, $dB \propto \dfrac{1}{r^2}$

Combining these:

$$dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\,\sin\theta}{r^2}, \qquad \text{or in vector form} \qquad d\vec B = \frac{\mu_0}{4\pi}\,\frac{I\,(d\vec l \times \hat r)}{r^2}$$

where $\mu_0 = 4\pi \times 10^{-7}$ T m A$^{-1}$ is the permeability of free space. The direction of $d\vec B$ is perpendicular to both $d\vec l$ and $\vec r$ (right-hand screw rule). The field is maximum when $\theta = 90^\circ$ and zero when $\theta = 0^\circ$ (a point along the line of the element).

(b) Field at the centre of a circular coil

Setup: A circular coil of $N$ turns and radius $r$ carries a current $I$. Let O be its centre.

            <-  dl
        .  -  ●  -  .
     .'       |       '.
    /         | r       \
   |          O  (dB out |
    \        of page)   /
     '.               .'
        '  -  .  -  '

1. Take a small element $d\vec l$ of the coil. The line from the element to O is a radius of length $r$, so the angle between $d\vec l$ and $\vec r$ is $\theta = 90^\circ$.

2. By the Biot–Savart law, the field at O due to this element is

$$dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\,\sin 90^\circ}{r^2} = \frac{\mu_0}{4\pi}\,\frac{I\,dl}{r^2}$$

3. For every element of the coil, $d\vec B$ at O points the same way: perpendicular to the plane of the coil, along its axis. So the contributions simply add:

$$B = \sum dB = \frac{\mu_0 I}{4\pi r^2}\sum dl$$

4. For $N$ turns, $\sum dl = N \times 2\pi r$:

$$B = \frac{\mu_0 I}{4\pi r^2} \times 2\pi r N$$ $$\boxed{B = \frac{\mu_0 N I}{2r}}$$

For a single turn, $B = \dfrac{\mu_0 I}{2r}$. The direction is given by the right-hand thumb rule: curl the fingers along the current, and the thumb points along $B$.

(c) Numerical: field on the axis

Given:

• number of turns, $N = 100$
• radius, $R = 10$ cm $= 0.10$ m
• current, $I = 5$ A
• distance from the centre, $x = 5$ cm $= 0.05$ m

Field on the axis of a circular coil:

$$B = \frac{\mu_0 N I R^2}{2\,(R^2 + x^2)^{3/2}}$$

Here $R^2 + x^2 = 0.01 + 0.0025 = 0.0125\ \text{m}^2$, so $(R^2 + x^2)^{3/2} = 0.0125 \times \sqrt{0.0125} = 1.398 \times 10^{-3}\ \text{m}^3$. Then

$$B = \frac{4\pi \times 10^{-7} \times 100 \times 5 \times 0.01}{2 \times 1.398 \times 10^{-3}} = \frac{6.283 \times 10^{-6}}{2.795 \times 10^{-3}}$$ $$B \approx 2.25 \times 10^{-3}\ \text{T}$$

The magnetic field on the axis is about $2.25 \times 10^{-3}$ T, directed along the axis. (Check: at the centre it would be $\mu_0 NI/2R = 3.14 \times 10^{-3}$ T, and at this point it is smaller by the factor $0.8^{3/2} = 0.716$.)

a) i) Magnitude of the force

Setup: Two long straight parallel conductors X and Y are a distance $r$ apart in air and carry currents $I_1$ and $I_2$ in the same direction.

        X            Y
        ^ I1         ^ I2
        |            |
        |  --> F <-- |      (attraction)
        |            |
        |<---- r --->|

1. Field of X at Y: The field produced by the long wire X at a distance $r$ is

$$B_1 = \frac{\mu_0 I_1}{2\pi r}$$

It is perpendicular to Y (into the page at Y, by the right-hand grip rule).

2. Force on a length $l$ of Y: Y carries $I_2$ perpendicular to $B_1$, so

$$F = B_1 I_2 l \sin 90^\circ = \frac{\mu_0 I_1 I_2\, l}{2\pi r}$$

3. Force per unit length:

$$\boxed{\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi r}}$$

By the same argument, X experiences an equal and opposite force (Newton's third law).

4. Nature (see part ii): Fleming's left-hand rule shows that the force on Y points towards X, and the force on X points towards Y. So the force is attractive when the currents are in the same direction, and repulsive when they are in opposite directions.

a) ii) Nature of the force

Both currents flow upward, i.e. in the same direction, so the force is attractive: the conductors pull towards each other.

a) iii) Definition of one ampere

One ampere: Put $I_1 = I_2 = 1$ A and $r = 1$ m in $\dfrac{F}{l} = \dfrac{\mu_0 I_1 I_2}{2\pi r}$:

$$\frac{F}{l} = \frac{4\pi \times 10^{-7} \times 1 \times 1}{2\pi \times 1} = 2 \times 10^{-7}\ \text{N m}^{-1}$$

So one ampere is the constant current which, when flowing through each of two infinitely long, straight, parallel conductors of negligible cross-section placed 1 m apart in vacuum, produces a force of $2 \times 10^{-7}$ N per metre of length between them.

b) i) Force on side AB

Given:

• $B = 4.5 \times 10^{-3}$ T, horizontal, to the right
• $I = 2.5$ A, flowing down from B to A
• length of AB, $l = 7$ cm $= 0.07$ m
• AB is vertical, so it is perpendicular to the field: $\theta = 90^\circ$

$$F = BIl\sin\theta = 4.5 \times 10^{-3} \times 2.5 \times 0.07 \times 1 = 7.9 \times 10^{-4}\ \text{N}$$

The force on AB is $7.9 \times 10^{-4}$ N (more exactly $7.875 \times 10^{-4}$ N). By Fleming's left-hand rule (current downward, field to the right), it acts perpendicular to the plane of the frame, out of the page. Side CD carries current upward, so it feels an equal force into the page. The two forces form a couple that tends to rotate the frame.

b) ii) Why side BC experiences no force

The force on a current-carrying conductor is $F = BIl\sin\theta$, where $\theta$ is the angle between the conductor and the field. Side BC is parallel to the magnetic field (its current flows from C to B, exactly opposite to $\vec B$), so $\theta = 180^\circ$ and $\sin\theta = 0$:

$$F = BIl\sin 180^\circ = 0$$

The moving charges in BC travel along the field lines, so $\vec v \times \vec B = 0$ and they feel no magnetic force.

a) Field due to a long straight conductor (Ampere's circuital law)

Ampere's circuital law: The line integral of the magnetic field $\vec B$ around any closed path equals $\mu_0$ times the net current enclosed by the path:

$$\oint \vec B \cdot d\vec l = \mu_0 I$$

              . - - .
           .'    B    '.
          /   ↺        \
         |       ⊙ I    |  <- circular Amperian loop, radius r
          \      |--r-->/
           '.         .'
              ' - - '

1. Take a long straight conductor carrying current $I$ (out of the page). To find the field at a point P at perpendicular distance $r$, draw a circular Amperian loop of radius $r$ centred on the wire and passing through P.

2. By symmetry, $B$ has the same magnitude at every point of this circle, and it is directed along the tangent to the circle. So $\vec B$ is parallel to $d\vec l$ everywhere ($\theta = 0^\circ$):

$$\oint \vec B \cdot d\vec l = \oint B\,dl\cos 0^\circ = B \oint dl = B\,(2\pi r)$$

3. The current enclosed is $I$, so by Ampere's law

$$B\,(2\pi r) = \mu_0 I$$ $$\boxed{B = \frac{\mu_0 I}{2\pi r}}$$

The field lines are concentric circles around the wire, and their direction is given by the right-hand grip rule. The magnetic field intensity is $H = \dfrac{B}{\mu_0} = \dfrac{I}{2\pi r}$.

b) Maximum and minimum force

Given: $l = 0.5$ m, $I = 2$ A, $B = 0.4$ T, and $F = BIl\sin\theta$.

• Maximum (wire perpendicular to the field, $\theta = 90^\circ$):

$$F_{max} = BIl = 0.4 \times 2 \times 0.5 = 0.4\ \text{N}$$

• Minimum (wire parallel to the field, $\theta = 0^\circ$ or $180^\circ$):

$$F_{min} = 0$$

i) Voltage sensitivity

The voltage sensitivity of a moving coil galvanometer is the deflection produced per unit potential difference applied across it, $\dfrac{\theta}{V}$. Its unit is rad V$^{-1}$ (or div V$^{-1}$).

At equilibrium, the deflecting torque equals the restoring torque: $NIAB = k\theta$. Since $I = V/R$,

$$\frac{\theta}{V} = \frac{NAB}{kR}$$

So the voltage sensitivity depends on:

• the number of turns $N$
• the area $A$ of the coil
• the magnetic flux density $B$ (it increases with each of these three)
• the torsional constant $k$ of the suspension or spring (it decreases as $k$ increases)
• the resistance $R$ of the coil (it decreases as $R$ increases)

ii) Why the field is radial

The magnet pole pieces are concave and there is a soft-iron core inside the coil, so the field is radial. The plane of the coil then always lies along the field lines whatever its deflection, so the torque is always maximum: $\tau = NIAB$ (with $\sin 90^\circ = 1$). Hence $k\theta = NIAB$, so $\theta \propto I$, and the galvanometer has a linear (uniform) scale.

a) Field at the centre of a circular coil (Biot–Savart law)

By the Biot–Savart law, the field due to a current element is $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$.

Setup: A circular coil of $N$ turns and radius $r$ carries a current $I$. Let O be its centre.

            <-  dl
        .  -  ●  -  .
     .'       |       '.
    /         | r       \
   |          O  (dB out |
    \        of page)   /
     '.               .'
        '  -  .  -  '

1. Take a small element $d\vec l$ of the coil. The line from the element to O is a radius of length $r$, so the angle between $d\vec l$ and $\vec r$ is $\theta = 90^\circ$.

2. By the Biot–Savart law, the field at O due to this element is

$$dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\,\sin 90^\circ}{r^2} = \frac{\mu_0}{4\pi}\,\frac{I\,dl}{r^2}$$

3. For every element of the coil, $d\vec B$ at O points the same way: perpendicular to the plane of the coil, along its axis. So the contributions simply add:

$$B = \sum dB = \frac{\mu_0 I}{4\pi r^2}\sum dl$$

4. For $N$ turns, $\sum dl = N \times 2\pi r$:

$$B = \frac{\mu_0 I}{4\pi r^2} \times 2\pi r N$$ $$\boxed{B = \frac{\mu_0 N I}{2r}}$$

For a single turn, $B = \dfrac{\mu_0 I}{2r}$. The direction is given by the right-hand thumb rule: curl the fingers along the current, and the thumb points along $B$.

b) Number of charge carriers per unit volume (Hall effect)

Given:

• thickness in the direction of $B$, $t = 2.5$ mm $= 2.5 \times 10^{-3}$ m
• current, $I = 150$ mA $= 0.150$ A
• flux density, $B = 0.5$ T
• Hall voltage, $V_H = 8.75$ mV $= 8.75 \times 10^{-3}$ V
• charge magnitude, $e = 1.6 \times 10^{-19}$ C

The Hall voltage is $V_H = \dfrac{BI}{nte}$, so

$$n = \frac{BI}{V_H\, t\, e} = \frac{0.5 \times 0.150}{8.75 \times 10^{-3} \times 2.5 \times 10^{-3} \times 1.6 \times 10^{-19}}$$ $$n = \frac{0.075}{3.5 \times 10^{-24}} = 2.1 \times 10^{22}\ \text{m}^{-3}$$

The number of free charge carriers is about $2.1 \times 10^{22}$ per m$^3$. This is much smaller than in metals (about $10^{28}$ m$^{-3}$), as expected for a semiconductor.

c) Force between two parallel conductors; one ampere

Setup: Two long straight parallel conductors X and Y are a distance $r$ apart in air and carry currents $I_1$ and $I_2$ in the same direction.

        X            Y
        ^ I1         ^ I2
        |            |
        |  --> F <-- |      (attraction)
        |            |
        |<---- r --->|

1. Field of X at Y: The field produced by the long wire X at a distance $r$ is

$$B_1 = \frac{\mu_0 I_1}{2\pi r}$$

It is perpendicular to Y (into the page at Y, by the right-hand grip rule).

2. Force on a length $l$ of Y: Y carries $I_2$ perpendicular to $B_1$, so

$$F = B_1 I_2 l \sin 90^\circ = \frac{\mu_0 I_1 I_2\, l}{2\pi r}$$

3. Force per unit length:

$$\boxed{\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi r}}$$

By the same argument, X experiences an equal and opposite force (Newton's third law).

4. Nature: Fleming's left-hand rule shows that the force on Y points towards X, and the force on X points towards Y. So the force is attractive when the currents are in the same direction, and repulsive when they are in opposite directions.

One ampere: Put $I_1 = I_2 = 1$ A and $r = 1$ m in $\dfrac{F}{l} = \dfrac{\mu_0 I_1 I_2}{2\pi r}$:

$$\frac{F}{l} = \frac{4\pi \times 10^{-7} \times 1 \times 1}{2\pi \times 1} = 2 \times 10^{-7}\ \text{N m}^{-1}$$

So one ampere is the constant current which, when flowing through each of two infinitely long, straight, parallel conductors of negligible cross-section placed 1 m apart in vacuum, produces a force of $2 \times 10^{-7}$ N per metre of length between them.

(B) $\dfrac{\mu_0 I}{2R}$

By the Biot–Savart law, every element $dl$ of the loop is at distance $R$ from the centre and makes $90^\circ$ with the radius, so

$$B = \frac{\mu_0 I}{4\pi R^2}\,(2\pi R) = \frac{\mu_0 I}{2R}$$

For $N$ turns this becomes $\mu_0 N I/2R$.

i) Force per unit length

Setup: Two long straight parallel conductors P and Q are a distance $r$ apart in air and carry currents $I_1$ and $I_2$ in the same direction.

        P            Q
        ^ I1         ^ I2
        |            |
        |  --> F <-- |      (attraction)
        |            |
        |<---- r --->|

1. Field of P at Q: The field produced by the long wire P at a distance $r$ is

$$B_1 = \frac{\mu_0 I_1}{2\pi r}$$

It is perpendicular to Q (into the page at Q, by the right-hand grip rule).

2. Force on a length $l$ of Q: Q carries $I_2$ perpendicular to $B_1$, so

$$F = B_1 I_2 l \sin 90^\circ = \frac{\mu_0 I_1 I_2\, l}{2\pi r}$$

3. Force per unit length:

$$\boxed{\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi r}}$$

By the same argument, P experiences an equal and opposite force (Newton's third law).

4. Nature: Fleming's left-hand rule shows that the force on Q points towards P, and the force on P points towards Q. So the force is attractive when the currents are in the same direction, and repulsive when they are in opposite directions.

ii) Nature of the force

The currents in P and Q are both upward, i.e. in the same direction, so the force is attractive: the wires pull towards each other.

iii) Distance halved

Since $\dfrac{F}{l} \propto \dfrac{1}{r}$, halving the distance ($r \to r/2$) gives

$$\frac{F'}{l} = \frac{\mu_0 I_1 I_2}{2\pi (r/2)} = 2\,\frac{F}{l}$$

The force per unit length becomes double (and is still attractive).

Ampere's circuital law: $\displaystyle\oint \vec B \cdot d\vec l = \mu_0 I$, where $I$ is the current enclosed by the closed path.

1. Consider a long straight conductor carrying current $I$. Choose a circular Amperian loop of radius $r$, centred on the wire, in a plane perpendicular to it.

2. By symmetry, $B$ has the same magnitude everywhere on the loop and is tangential to it, so $\vec B \parallel d\vec l$:

$$\oint \vec B \cdot d\vec l = B \oint dl = B\,(2\pi r)$$

3. Applying Ampere's law:

$$B\,(2\pi r) = \mu_0 I \quad\Rightarrow\quad \boxed{B = \frac{\mu_0 I}{2\pi r}}$$

The field lines are concentric circles around the wire, with direction given by the right-hand grip rule. $B \propto I$ and $B \propto \dfrac{1}{r}$.

a) Field at a point on the axis of a circular coil

              dl (out of page)
               ●  .
               |     ' .   s = √(R² + x²)
             R |         ' .        dB⊥ ↑
               |      φ     ' .  ↗ dB
               O----------------P -----> dB_x (along axis)
               |<------ x ----->|
               |
               ●  dl' (diametrically opposite)

Setup: A coil of radius $R$ carries a current $I$. P is a point on its axis at distance $x$ from the centre O.

1. Take an element $d\vec l$ at the top of the coil. Its distance from P is $s = \sqrt{R^2 + x^2}$, and $d\vec l \perp \vec s$. By the Biot–Savart law,

$$dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\sin 90^\circ}{s^2} = \frac{\mu_0}{4\pi}\,\frac{I\,dl}{R^2 + x^2}$$

$d\vec B$ is perpendicular to $\vec s$, in the plane containing $\vec s$ and the axis.

2. Resolve $dB$ into:

   • a component $dB\cos\phi$ perpendicular to the axis
   • a component $dB\sin\phi$ along the axis

   Here $\phi$ is the angle between $\vec s$ and the axis, so $\sin\phi = \dfrac{R}{\sqrt{R^2 + x^2}}$.

3. The diametrically opposite element $d\vec l'$ gives an equal perpendicular component in the opposite direction, so the perpendicular components cancel in pairs. The axial components all point the same way and add.

4. Total field:

$$B = \sum dB\sin\phi = \frac{\mu_0 I}{4\pi (R^2 + x^2)}\cdot\frac{R}{\sqrt{R^2 + x^2}}\sum dl$$

With $\sum dl = 2\pi R$:

$$B = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} \times 2\pi R$$ $$\boxed{B = \frac{\mu_0 I R^2}{2\,(R^2 + x^2)^{3/2}}}$$

It acts along the axis. For $N$ turns, multiply by $N$. At the centre ($x = 0$), this reduces to $B = \dfrac{\mu_0 I}{2R}$.

b) Why the Hall voltage is larger (more measurable) in semiconductors

The Hall voltage is

$$V_H = \frac{BI}{n e t}$$

so for the same $B$, $I$ and thickness $t$, $V_H \propto \dfrac{1}{n}$. A semiconductor has a much smaller number of charge carriers per unit volume ($n \sim 10^{20}$ to $10^{22}$ m$^{-3}$) than a metal ($n \sim 10^{28}$ m$^{-3}$). For the same current, the carriers in a semiconductor must therefore drift much faster ($I = nAev$), feel a bigger magnetic force $evB$, and build up a larger Hall field. So $V_H$ is millions of times bigger in a semiconductor and easy to measure, while in metals it is extremely small (of the order of microvolts).

c) Field lines for opposite currents

          ⊙ I (out of page)          ⊗ I (into page)

      .- - - - .      ↑   ↑      .- - - - .
    /  .- - -.   \    ↑   ↑    /  .- - -.   \
   |  |   ⊙   |   |   ↑   ↑   |  |   ⊗   |   |
    \  '- - -'   /    ↑   ↑    \  '- - -'   /
      '- - - - '      ↑   ↑      '- - - - '
     anticlockwise   crowded       clockwise
                    (strong field)

Each wire is surrounded by concentric circular field lines: anticlockwise around the wire with current out of the page, and clockwise around the wire with current into the page. Between the wires both fields point the same way (upward here), so the lines are crowded and the field is strong. Outside the wires the fields partly cancel, so the lines are spread out. The wires repel each other.

d) Force per unit length

Given: $r = 0.1$ m, $I_1 = 3$ A, $I_2 = 2$ A (same direction), $\mu_0 = 4\pi \times 10^{-7}$ H m$^{-1}$.

$$\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi r} = \frac{4\pi \times 10^{-7} \times 3 \times 2}{2\pi \times 0.1} = \frac{2 \times 10^{-7} \times 6}{0.1}$$ $$\frac{F}{l} = 1.2 \times 10^{-5}\ \text{N m}^{-1}$$

Each conductor experiences a force of $1.2 \times 10^{-5}$ N per metre, and the force is attractive because the currents are in the same direction.

(i) Biot-Savart law

The magnetic field $d\vec B$ at a point due to a small current element $I\,d\vec l$ is directly proportional to the current, the length $dl$ and $\sin\theta$ (the angle between $d\vec l$ and the line to the point), and inversely proportional to the square of the distance $r$:

$$dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\sin\theta}{r^2}$$

(ii) Field at the centre of a circular coil

Let the coil have radius $R$ and carry current $I$. Every element $dl$ is at distance $R$ from the centre and perpendicular to the radius, so $\theta = 90^\circ$. All elements give field in the same direction at the centre, so the fields add directly:

$$B = \int \frac{\mu_0}{4\pi}\frac{I\,dl}{R^2} = \frac{\mu_0 I}{4\pi R^2}\int dl = \frac{\mu_0 I}{4\pi R^2}(2\pi R)$$ $$B = \frac{\mu_0 I}{2R}$$

For a coil of $N$ turns, $B = \dfrac{\mu_0 N I}{2R}$.

(iii) Direction

The field at the centre is perpendicular to the plane of the coil, along its axis. It is given by the right-hand rule: curl the fingers of the right hand along the current, and the thumb points along the field (into the plane for clockwise current, out of the plane for anticlockwise current).

(a) Biot-Savart law: The magnetic field $dB$ at a point due to a small current element $I\,dl$ is directly proportional to $I$, $dl$ and $\sin\theta$ and inversely proportional to the square of the distance $r$:

$dB = \frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2}$

Here $\theta$ is the angle between $d\vec l$ and the line joining the element to the point.

Field of a long straight conductor: Take a circular Amperian loop of radius $r$ around the wire, in a plane perpendicular to it. By symmetry $B$ has the same magnitude at every point of the loop and is tangent to it. Ampere's circuital law gives

$\oint \vec B\cdot d\vec l = \mu_0 I \;\Rightarrow\; B(2\pi r) = \mu_0 I \;\Rightarrow\; B = \frac{\mu_0 I}{2\pi r}$

(b) The current of 2 A splits equally between the two paths abc and adc, which have equal resistance, so each path carries $1$ A. Take the field as into the page, with a at the bottom-left, b at the bottom-right, c at the top-right and d at the top-left.

Each side has $F = BIl = 0.25\times1\times0.20 = 0.05$ N.

Every side feels 0.05 N. The forces are perpendicular to each side, and they reverse if the field is taken out of the page. The net force is $0.05\sqrt2\times 2 = 0.14$ N, which equals the force on a straight wire ac carrying 2 A.

(a) One ampere

One ampere is the constant current which, when flowing in each of two infinitely long, straight, parallel conductors of negligible cross-section placed 1 m apart in vacuum, produces a force of $2 \times 10^{-7}$ N per metre of length between them.

(b) Contraction of a solenoid

In a solenoid, neighbouring turns are close together and parallel, and they all carry current in the same direction. Parallel conductors with currents in the same direction attract each other ($F/l = \mu_0 I_1 I_2/2\pi r$). So every turn is pulled towards its neighbours and the whole coil is squeezed along its length: the solenoid contracts when a current is passed. The effect is easy to see in a loosely wound, flexible spring-like coil.

(c)(i) Current in the conductor

Given:

• mass per unit length, $\lambda = 0.2$ g m$^{-1}$ $= 2 \times 10^{-4}$ kg m$^{-1}$
• $B = 1$ T (into the page, perpendicular to the rod)
• $g = 9.8$ m s$^{-2}$

The tension in the wires is zero, so the upward magnetic force on the rod balances its weight:

$$BIl = mg = \lambda l g \quad\Rightarrow\quad I = \frac{\lambda g}{B} = \frac{2 \times 10^{-4} \times 9.8}{1}$$ $$I = 1.96 \times 10^{-3}\ \text{A} \approx 2.0\ \text{mA}$$

By Fleming's left-hand rule (field into the page, force upward), the current in the rod must flow from left to right.

(c)(ii) Torque on the coil

Given:

• $N = 100$
• $I = 1.96 \times 10^{-3}$ A
• $A = 100$ cm$^2$ $= 1.0 \times 10^{-2}$ m$^2$
• $B = 10$ T

The plane of the coil is parallel to the field, so its axis (normal) makes $\theta = 90^\circ$ with $\vec B$ and the torque is maximum:

$$\tau = NIAB\sin 90^\circ = 100 \times 1.96 \times 10^{-3} \times 1.0 \times 10^{-2} \times 10$$ $$\tau = 1.96 \times 10^{-2}\ \text{N m} \approx 0.020\ \text{N m}$$

The torque produced is about $1.96 \times 10^{-2}$ N m.

(i) Magnetic flux density

The magnetic flux density $B$ at a point is the force per unit current per unit length on a straight conductor placed perpendicular to the field at that point:

$$B = \frac{F}{Il}$$

Its SI unit is the tesla (T), where 1 T = 1 N A$^{-1}$ m$^{-1}$. Equivalently, it is the magnetic flux per unit area normal to the field, $B = \phi/A$.

(ii) Why a Hall probe is not suitable

The gap between the two magnets on the yoke is very narrow, and a Hall probe (with its holder) is too thick to fit properly into it with its face perpendicular to the field. Also, a Hall probe has to be calibrated against a known field before it can give $B$, and in such a weak field (0.02 to 0.04 T) the Hall voltage is small, which makes an accurate reading difficult.

(iii) Measuring $B$ with a current-carrying wire (current balance method)

          clamp
            |
   ---------+---------------  stiff straight wire (length L in the gap),
            |  [ N ]  [ S ]   perpendicular to the field, not touching
            |   yoke + magnets
     +------------------------+
     |   top-pan balance  0.00 g |
     +------------------------+
   wire in series with: battery -- rheostat -- ammeter (A)

1. Place the steel yoke with its magnets on an electronic top-pan balance. Record (or tare to zero) the reading with no current.
2. Clamp a stiff straight wire horizontally between the poles, perpendicular to the magnetic field, so that it does not touch the magnets or the yoke. Connect it in series with a battery, a rheostat and an ammeter.
3. Switch on a current $I$, read on the ammeter. The field exerts a vertical force $F = BIL$ on the wire, where $L$ is the length of wire inside the field. By Newton's third law, an equal and opposite force acts on the magnets, so the balance reading changes by $\Delta m$:

$$F = (\Delta m)\,g = BIL$$

4. Measure $L$ (the length of the pole faces along the wire) with a ruler. Then

$$B = \frac{(\Delta m)\,g}{IL}$$

5. For better accuracy, repeat for several currents (adjusted with the rheostat). Plot $\Delta m$ against $I$: this is a straight line through the origin with gradient $\dfrac{BL}{g}$, so

$$B = \frac{g \times \text{gradient}}{L}$$

Reverse the current to check that the change in reading is equal and opposite.

For $B \approx 0.03$ T, $I = 5$ A and $L = 5$ cm, $F \approx 7.5 \times 10^{-3}$ N, i.e. $\Delta m \approx 0.77$ g, which is easily measured on a balance reading to 0.01 g.