Chapter 12: Magnetic Effects of Electric Current

The essential facts we use are

• Every compass needle is itself a small, light magnetic dipole. Its north pole is conventionally painted red and can turn freely about a pivot.
• A bar magnet produces a magnetic field in the space surrounding it. At each point this field is represented by a vector $$\vec{B}$$ whose direction is tangent to the magnetic field line through that point.
• Whenever a magnetic dipole of moment $$\vec{\mu}$$ is placed in an external magnetic field $$\vec{B}$$, it experiences a torque $$\vec{\tau}=\vec{\mu}\times\vec{B}$$ which tends to rotate the dipole so that its magnetic moment lines up with the field.

Now let the compass be brought close to one pole of the bar magnet.

1. The bar magnet’s field $$\vec{B}$$ at the compass position is not the same as the Earth’s field. Its magnitude is much larger and its direction is nearly along the lines that emerge from (or enter into) the bar magnet.
2. The compass needle, being a dipole, experiences the torque $$\vec{\tau}=\vec{\mu}\times\vec{B}_{\text{bar-magnet}}$$.
3. This torque is zero only when $$\vec{\mu}$$ is parallel or antiparallel to $$\vec{B}_{\text{bar-magnet}}$$. Starting from its original orientation (along the Earth’s field), the needle therefore begins to turn until it reaches the nearest zero-torque position, i.e. until it aligns with the local field lines of the bar magnet.

Because the direction of the bar magnet’s field is different from that of the Earth’s magnetic field, the alignment process is observed as a deflection of the compass needle.

Thus the compass needle is deflected whenever it is brought near a bar magnet, simply because the magnet’s own magnetic field exerts a torque on the magnetic dipole of the needle and forces it to turn.

Given data

• The power line is straight and horizontal.
• The electric current $$I$$ flows from east to west.
• We have to find the direction of the magnetic field $$\vec B$$
• (i) at a point directly below the wire    (ii) at a point directly above the wire.

Principle used — Right-Hand Thumb Rule

Hold the conductor with the right hand so that the thumb points in the direction of the current. The curl of the fingers then gives the direction of the concentric magnetic field lines surrounding the wire.

Step 1 : Fix the thumb along the current

Current is from east → west, i.e. from right to left when one faces north. Point the right-hand thumb towards the west.

Step 2 : Trace the finger-curl

With the thumb fixed toward the west you will notice:

• Above the conductor your curled fingers point towards the north.
• Below the conductor your curled fingers point towards the south.

Step 3 : (Optional check by vector product)

Take east as the +x direction, north as +y and vertically up as +z. The current vector is $$\vec I = -I\,\hat i$$. For a point on the z-axis $$\vec r = (0,0,\pm r)$$. The magnetic field of a long straight wire is along $$\vec I \times \vec r$$.

Above the wire ( $$\vec r = (0,0,+r)$$ ): $$\vec B \propto \vec I \times \vec r = (-I\,\hat i) \times ( +r\,\hat k) = +Ir\,\hat j$$, i.e. towards north.

Below the wire ( $$\vec r = (0,0,-r)$$ ): $$\vec B \propto (-I\,\hat i) \times ( -r\,\hat k) = -Ir\,\hat j$$, i.e. towards south.

Conclusion

• The magnetic field at a point directly below the power line is towards the south.
• The magnetic field at a point directly above the power line is towards the north.

What is to be drawn ?

Magnetic field lines are imaginary curves whose tangent at any point gives the direction of the magnetic field $$\vec B$$ at that point. For a bar magnet the pattern is characteristic and must obey the three standard properties:

• Each line forms a closed loop: outside the magnet it goes from the north pole (N) to the south pole (S); inside the magnet it continues from S to N.
• No two field-lines ever intersect.
• The crowding of the lines represents field strength; they are densest near the poles.

Experiment — plotting the field with a compass

1. Place a white sheet of paper on a board and keep the bar magnet roughly at its centre. Mark the positions of the N and S poles.
2. Put a small plotting compass near the N-pole. Its needle sets itself along the field. Put a dot at the centre of the compass at each end of the needle.
3. Shift the compass so that the south-end of the needle now sits on the second dot. Again mark a new dot at the north-end of the needle.
4. Continue step 3 until the path loops round to the S-pole of the magnet. Join the successive dots with a smooth curve; this is one field line.
5. Repeat the entire procedure starting from several different points around the N-pole to obtain a complete family of lines.

What the final diagram must show

• Outside the magnet, smooth curves emerge normally from the N-pole, arch through space, and re-enter the S-pole.
• Inside the magnet the lines run straight from S to N, thereby completing the closed loops.
• Lines are closer together near N and S, indicating a stronger field there.

Written description if a diagram cannot be supplied in the answer sheet

“Draw several smooth curves starting from the north pole of the bar magnet, bulging outward, and ending at the south pole. Inside the magnet draw the curves straight from S to N so that each curve is a closed loop. Keep the curves denser near both poles and ensure no two curves cross.”

Concept of field lines

To visualise a vector field such as a magnetic field $$\vec B$$ we draw a family of imaginary curves, called magnetic field lines, whose geometry and density reproduce the direction and relative magnitude of $$\vec B$$ in every region of space.

Experimentally they can be traced, for instance, by the orientation of many small compass needles or by the pattern taken up by iron filings placed around a magnet.

Properties with reasons

1. Closed continuous curves – Outside a magnet the lines emerge from the North pole and enter the South pole; inside the magnet they continue from the South to the North pole. Hence each line forms one unbroken closed loop, reflecting the fact that $$\nabla\!\cdot\!\vec B = 0$$ (no magnetic monopoles are known).

2. Direction of the field – The tangent drawn to a field line at any point gives the direction of $$\vec B$$ at that point. A tiny compass needle aligns itself along the tangent.

3. Relative strength of the field – The density of lines is taken proportional to the magnitude of the field: the closer the lines, the stronger the field, i.e. $$|\vec B| \propto n/A$$ where $$n$$ is the number of lines cutting a small area $$A$$ held perpendicular to them.

4. No two lines ever intersect – If they did, two different tangents (and therefore two different directions of $$\vec B$$) would exist at the point of intersection, which is impossible because the magnetic field has a single well-defined direction at every point.

5. Field lines show lateral repulsion – In diagrams the lines spread out sideways as though they repel one another. This graphical convention emphasises that the field prefers to occupy all of space rather than crowd into regions already having strong field.

6. They never start or end in empty space – While electric field lines can terminate on charges, magnetic field lines can only close upon themselves; there are no isolated North or South magnetic charges.

These six statements summarise every essential geometrical and physical feature of magnetic field lines required at Class 10 level.

Magnetic field lines are an aid for picturing the vector magnetic field $$\vec B$$. At every point on a field line, the tangent to that line is, by definition, parallel to $$\vec B$$ at that point.

Assume, just for contradiction, that two lines L₁ and L₂ intersect at a point P. Draw the two curves so that they cross once at P; mark the tangents T₁ and T₂ to the curves at P.

Because the curves are distinct, their tangents are not collinear: $$T_1 \neq T_2$$. Therefore the construction would assign two different directions to the same magnetic field vector at point P – one along T₁, the other along T₂.

But the magnetic field at any single point has one, and only one, well-defined direction determined by the vector sum of all sources:$$\vec B(P)=\vec B_1(P)+\vec B_2(P)+\cdots$$Hence a point cannot simultaneously possess two directions for $$\vec B$$.

The assumption that lines L₁ and L₂ intersect therefore contradicts the uniqueness of the magnetic field direction. Consequently, two magnetic field lines never intersect.

Step 1 : Visualise the situation

The conducting loop lies flat on the table (the loop is in the plane of the page if we look at the table from above). The current is given to flow clockwise when viewed from above.

Step 2 : Recall the right-hand rule for a current loop

Grasp the loop with your right hand so that the curled fingers point in the direction of the current at every point of the loop. Your extended thumb then points along the direction of the magnetic field at the centre of the loop (i.e. “inside” the loop). The field outside the loop is in the opposite direction because magnetic field lines form closed curves.

Step 3 : Apply the rule to the given current direction

• Place the right hand above the table and curl the fingers clockwise (matching the current).
• The thumb now points downwards, i.e. into the table (into the plane of the page).

Step 4 : State the directions

• Inside the loop (central region) the magnetic field is directed into the table (perpendicular to the plane, away from the observer).
• Outside the loop the field lines emerge from the table, so the magnetic field is directed out of the table (towards the observer).

Diagram to draw (for the student): Sketch a circle representing the loop, mark clockwise arrows for current, place several ✕ symbols inside the circle to denote field going into the page, and several • symbols outside to denote field coming out of the page.

For a magnetic field to be uniform, the magnetic induction $$\vec{B}$$ must have the same magnitude and the same direction at every point in the region. In diagrams, magnetic field lines are drawn so that

• the spacing between adjacent lines indicates the strength of the field (closer → stronger);
• arrow-heads on the lines give the direction of $$\vec{B}$$.

Therefore, a uniform magnetic field is shown by a set of straight, parallel, equally-spaced lines all carrying arrow-heads in the same direction.

How to draw it:

1. Draw 5 – 6 straight horizontal (or vertical) lines that are perfectly parallel to each other.
2. Keep the gap between any two neighbouring lines the same throughout.
3. Put arrow-heads on every line pointing, say, from left to right (or bottom to top). All arrows must point the same way.

This picture conveys that at every point in the shaded region the magnetic field vector $$\vec{B}$$ has

• identical magnitude (equal spacing), and
• identical direction (parallel lines with identical arrow-heads).

Such a diagram is the standard representation of a uniform magnetic field in physics.

The strength of the magnetic field $$B$$ inside a long, straight, closely wound solenoid can be found from Ampere’s circuital law.

Consider a rectangular amperian loop that runs partly through the interior of the solenoid (length $$\\ell$$) and partly outside (where the field is negligible). For this loop

$$\oint \vec{B}\cdot d\vec{l}=B\,\ell$$

The line integral equals $$\mu_0$$ times the net current enclosed by the loop. If the solenoid has $$n$$ turns per unit length and carries current $$I$$, the enclosed current equals $$n\ell I$$. Therefore,

$$B\,\ell = \mu_0 n\ell I \;\;\Rightarrow\;\; B = \mu_0 n I$$

Important points:

• $$\mu_0$$ (permeability of free space), $$n$$ (turns per unit length) and $$I$$ (current) are constants for a given solenoid in a given experiment.
• The expression $$B = \mu_0 n I$$ contains no term that depends on the position inside the solenoid (provided we stay well away from the ends).

Hence the magnetic field is uniform—the same at every point inside a sufficiently long solenoid.

Therefore, the correct option is:

(d) is the same at all points.

Step 1  Interpret the directions shown in Fig. 12.14
In the textbook diagram the magnetic field $$\vec B$$ is represented by dots ( • ) coming out of the page, while the velocity $$\vec v$$ of the electron is upward along the page (towards the top edge). Thus, in Cartesian symbols
$$\vec v = v\,\hat{j},\qquad \vec B = B\,\hat{k}$$

Step 2  Use the force formula on a moving charge
The magnetic force on a charge $$q$$ moving with velocity $$\vec v$$ in a magnetic field $$\vec B$$ is
$$\vec F = q\,(\vec v \times \vec B).$$

Step 3  Evaluate the cross-product
$$\vec v \times \vec B = (v\,\hat{j}) \times (B\,\hat{k}) = vB\,(\hat{j} \times \hat{k}) = vB\,\hat{i}.$$
This result points towards the right side of the page ( +$$\hat{i}$$ direction).

Step 4  Insert the charge of an electron
For an electron, $$q = -e$$ (negative). Hence
$$\vec F = (-e)\,(vB\,\hat{i}) = -e vB\,\hat{i},$$
which is along −$$\hat{i}$$, that is, towards the left side of the page.

Step 5  Choose the correct option
The magnetic force acts to the left, so the correct choice is (b).

Given: A proton is allowed to move freely in a magnetic field (no electric field present).

Magnetic force on a moving charge
When a charge $$q$$ with velocity $$\vec v$$ enters a magnetic field $$\vec B$$, the magnetic (Lorentz) force on it is

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

The vector product $$\vec v \times \vec B$$ is always perpendicular to $$\vec v$$. Therefore,

1. $$\vec F \perp \vec v$$ at every instant.
2. The work done per unit time (power) by the magnetic force is

$$P = \vec F \cdot \vec v = 0,$$

because the dot-product of two perpendicular vectors is zero.

Consequences

• Kinetic energy $$\bigl(\tfrac12 m v^2\bigr)$$ is unchanged: the force does no work.
• Thus the speed (magnitude of velocity) $$v$$ remains constant.
• The direction of $$\vec v$$ continuously changes (the path is a circle or helix), so the velocity vector changes.
• Momentum is $$\vec p = m\vec v.$$(Intrinsic) mass $$m$$ stays constant, but since $$\vec v$$ changes direction, $$\vec p$$ also changes.

Property–wise verdict

Hence only options (c) and (d) can change.

Known principle

A straight conductor carrying current $$I$$ and placed in a magnetic field of induction $$\vec{B}$$ experiences a force

$$ \vec{F}=I (\vec{L}\times\vec{B}) $$

When the conductor is perpendicular to the field (as in Activity 12.7), $$\theta = 90^{\circ}$$, so

$$ F = I L B \sin 90^{\circ}=I L B $$

The mechanical displacement of the rod AB is produced by this force; the larger the force, the greater the displacement. Hence we simply have to see how the product $$I L B$$ changes in each situation.

(i) Current in AB increased

• In the formula, $$F \propto I$$.
• If $$I$$ is made larger, $$F$$ increases in the same proportion.
• A larger force pushes the rod further, so the displacement increases.

(ii) A stronger horse-shoe magnet used

• A stronger magnet means a larger magnetic field magnitude $$B$$ between the poles.
• Since $$F \propto B$$, the force increases.
• Consequently, the rod is displaced through a greater distance.

(iii) Length of rod AB increased

• Only the portion of the rod lying inside the field contributes; making this length $$L$$ bigger makes $$F \propto L$$ larger.
• The increased force again produces a larger displacement of the rod.

Conclusion

In every case—higher current, stronger magnet, or longer rod—the magnetic force acting on AB becomes larger, so the rod is pushed farther from its initial position.

Given data

• The charged particle is an alpha-particle; hence its charge $$q$$ is positive.
• Initial velocity $$\vec v$$: towards West.
• Magnetic force (deflection) $$\vec F$$: towards North.

Principle used

For a charged particle moving in a magnetic field, the force is given by the vector product

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

Because $$q>0$$ (positive charge), the direction of $$\vec F$$ is the same as that of $$\vec v \times \vec B$$. We therefore have to find a vector $$\vec B$$ such that

$$\vec v \times \vec B = \text{towards North.}$$

Choosing a coordinate frame

Consequently,

• Towards West  $$\bigl(\text{opposite of East}\bigr) \;\Rightarrow\; \vec v = -\hat\imath.$$
• Towards North  $$\Rightarrow\; \vec F = +\hat\jmath.$$

Determining \(\vec B\)

Let us test the upward direction, $$\vec B = +\hat k$$ (along $$+z$$):

$$\vec v \times \vec B = \bigl(-\hat\imath\bigr) \times \hat k.$$

Recall the right-hand rule (or the standard order of cross products):

$$\hat\imath \times \hat k = -\hat\jmath \;\Rightarrow\; \bigl(-\hat\imath\bigr) \times \hat k = +\hat\jmath.$$

Thus, with $$\vec B$$ pointing upward, $$\vec v \times \vec B$$ is indeed towards $$+y$$ (North), exactly the required force direction.

Conclusion

The magnetic field must be directed upward.

Hence, the correct option is (d).

Step 1 · Identify the requirement
Domestic wiring has to protect (a) the appliances and (b) the user from accidental overheating or electric shock.

Step 2 · List safety devices / techniques

• Fuse (or MCB)
  A fuse is a short piece of wire with a low melting point connected in series with the live conductor. If the current exceeds the rated value, the wire melts and breaks the circuit, thereby preventing $$I^2R$$ heating that could cause fire. Modern houses often use an MCB (Miniature Circuit Breaker) that opens the circuit electromagnetically but serves the same purpose.
• Earthing (Earth wire)
  Every metal-cased appliance is connected to the earth pin of a three-pin plug and, through it, to a thick copper strip buried deep in moist soil. If the live wire accidentally touches the metal body, the leakage current is diverted to the ground through the low-resistance earth wire, keeping the potential of the appliance near $$0\;\text{V}$$ and protecting the user from shock.

Conclusion
The two common safety measures are therefore (i) use of a fuse/MCB and (ii) proper earthing of circuits and appliances.

Given data

• Power rating of the oven, $$P = 2\,\text{kW} = 2000\,\text{W}$$
• Mains voltage, $$V = 220\,\text{V}$$
• Current rating of the domestic circuit, $$I_{\text{max}} = 5\,\text{A}$$

Step 1 : Calculate the current required by the oven

The relation between power, voltage and current is

$$P = V I$$

Therefore, the current drawn by the oven is

$$I = \frac{P}{V}$$

Substituting the given values :

$$I = \frac{2000\,\text{W}}{220\,\text{V}}$$

$$I = 9.09\,\text{A}\;(\text{approximately})$$

Step 2 : Compare with the circuit rating

The oven needs about $$9.1\,\text{A}$$, but the circuit can safely supply only up to $$5\,\text{A}$$.

Since $$9.1\,\text{A} > 5\,\text{A}$$, operating the oven will cause an over-current.

Step 3 : Expected result

• The fuse (or MCB) in the domestic circuit will blow/trip to protect the wiring.
• As a consequence, the oven will switch off and the circuit will be interrupted.

Conclusion

Trying to run a 2 kW oven on a 5 A, 220 V circuit will overload the circuit and the protective fuse will melt (or the MCB will trip) immediately.

Concept of over-loading
Over-loading occurs when the current drawn by the appliances connected in a branch becomes larger than the current rating of that branch. If the supply voltage of the mains is $$V = 220\;\text{V}$$ and we suddenly plug in several devices whose total power consumption is $$P_{\text{tot}}$$, the current through the branch becomes

$$I = \frac{P_{\text{tot}}}{V}\;.$$

If this current $$I$$ exceeds the current rating $$I_{\max}$$ (fixed by the fuse or by the safe current-carrying capacity of the wires), the wires over-heat, the insulation may melt and a fire hazard is created. Hence we must limit the total current drawn from any one circuit.

Precautions to avoid over-loading

• Do not connect too many high-power appliances (iron, heater, geyser, microwave, etc.) to a single socket or to a single branch circuit at the same time, because this raises $$P_{\text{tot}}$$ and hence $$I$$ beyond the safe value.
• Use fuses or miniature circuit breakers (MCBs) of the correct current rating in every circuit so that the circuit is automatically opened if $$I$$ accidentally exceeds $$I_{\max}$$.
• Use wires of proper thickness (low resistance) so that they can safely carry the rated current without excessive heating.

If these simple precautions are followed, the current in any domestic circuit remains within safe limits and over-loading is avoided.

Step 1 : Recall the experimental fact
Following Oersted’s discovery, a straight conductor carrying current was placed vertically through a horizontal cardboard sheet sprinkled with iron filings. When current flowed, the filings arranged themselves in closed circular patterns around the wire. This shows that magnetic field lines form closed concentric circles centred on the wire.

Step 2 : Apply the Right-Hand Thumb Rule
Point the right-hand thumb in the direction of conventional current $$I$$ (from positive to negative terminal). The curled fingers then point along the direction of the magnetic field lines $$\vec B$$. Because the fingers wrap round the thumb, the field at every point traces a circle around the wire; each circle has the wire as its centre and lies in a plane perpendicular to the wire.

Step 3 : Check the options

• (a) Straight lines perpendicular to the wire  →  Incorrect; field lines are not straight.
• (b) Straight lines parallel to the wire  →  Incorrect; experiment shows circular pattern.
• (c) Radial lines originating from the wire  →  Incorrect; field lines do not radiate outward.
• (d) Concentric circles centred on the wire  →  Correct; matches both experiment and the right-hand thumb rule.

Conclusion
The magnetic field near a long straight current-carrying wire consists of concentric circles centred on the wire.

Concept recalled
Ohm’s Law gives the relation between current $$I$$, potential difference $$V$$ and resistance $$R$$ of a closed conducting path:

$$I = \frac{V}{R}$$

What happens in a short-circuit?
A short circuit occurs when the live wire and the neutral wire (or the positive and negative terminals) come into direct contact, providing an almost resistance-free path. In that situation:

• The resistance of the path falls to a value that is very close to $$0\,\Omega$$ (although never exactly zero in practice).

Effect on current
Substituting a very small resistance $$R_{\text{short}} \approx 0$$ in Ohm’s Law:

$$I_{\text{short}} = \frac{V}{R_{\text{short}}} \;\Rightarrow\; I_{\text{short}} \text{ becomes a very large number}$$.

Thus, at the moment of short circuit, the current in the circuit rises sharply to a very high value.

Choosing the correct option

1. Current reduces substantially – incorrect.
2. Current does not change – incorrect.
3. Current increases heavily – correct.
4. Current varies continuously – not the characteristic feature.

Therefore, the right choice is option (c).

The earth and almost every magnet we come across tell us that a magnetic field can exist even without electricity; Oersted’s experiment, on the other hand, showed that an electric current can also create a magnetic field. Thus, at Class 10 level we recognise two independent ways of setting up a magnetic field around a region.

1. By a permanent magnet (magnetisation of matter)
   

   A bar magnet, a horseshoe magnet or even the earth itself produces a magnetic field by virtue of the alignment of atomic magnetic moments inside the material. Outside the magnet the field lines start from the north pole and return to the south pole, forming closed loops. The strength and shape of the field depend on the material and its geometry; no electric current is required once the magnet has been made.

2. By passing an electric current through a conductor (electromagnetism)
   • Straight current‐carrying wire: Oersted found that a compass needle placed near a straight wire deflects when the wire carries current. For a long straight wire the magnitude of the magnetic field at a perpendicular distance $$r$$ is given by $$B = \frac{\mu_0 I}{2\pi r}$$, where $$I$$ is the current and $$\mu_0$$ the permeability of free space.
   • Circular coil or solenoid: If the same wire is bent into $$n$$ turns of radius $$R$$, the field at the centre becomes $$B = \frac{\mu_0 n I}{2R}$$. In a long solenoid of $$n$$ turns per unit length carrying current $$I$$, the interior field is nearly uniform and equals $$B = \mu_0 n I$$. Insertion of a soft‐iron core multiplies the field tremendously, giving a powerful electromagnet.

Therefore, the two basic methods are (i) using permanent magnets and (ii) using electric current through a conductor (i.e. an electromagnet).

For a straight conductor of length L carrying a current I in a uniform magnetic field of induction B, the magnetic force is given by the vector product

$$\vec F = I\,(\vec L \times \vec B)$$

Hence the magnitude of the force is

$$F = I L B \sin \theta$$

where $$\theta$$ is the angle between the direction of current (along $$\vec L$$) and the magnetic field $$\vec B$$.

1. The quantities $$I$$, $$L$$ and $$B$$ are fixed for the given conductor and field.
2. The factor that can vary is $$\sin \theta$$, whose maximum possible value is 1.

Therefore,

$$F_{\max}=I L B \times 1$$

when $$\sin \theta=1 \;\Longrightarrow\; \theta=90^{\circ}$$.

Thus, the force is largest when the conductor is oriented perpendicular to the direction of the magnetic field lines.

Given

• You sit with your back to the rear wall and face the front wall.
• An electron beam travels horizontally from the rear (back) wall towards the front wall.
• The beam is deflected towards your right side.

Step 1 – Choose axes

• Facing direction (front wall)  →  $$+z$$-axis.
• Your right side  →  $$+x$$-axis.
• Upward (towards the ceiling)  →  $$+y$$-axis.

Step 2 – Velocity of the electrons

Electrons move from back to front, so $$\vec v_{\mathrm e}=+v\,\hat z$$.

Step 3 – Magnetic force on an electron

The Lorentz force is $$\vec F=q\,\vec v\times\vec B$$. For an electron, $$q=-e$$, hence

$$\vec F_{\mathrm e}=(-e)(\vec v_{\mathrm e}\times\vec B).$$

The observed deflection is to the right: $$\vec F_{\mathrm e}=+F\,\hat x$$.

Step 4 – Determine $$\vec B$$

Substituting directions:

$$+F\,\hat x = (-e)(+v\,\hat z \times \vec B).$$

Divide by the negative constant $$-ev$$:

$$\hat z \times \vec B = -\hat x.$$

Using the right-hand rule, $$\hat z \times \hat y = -\hat x$$, so

$$\vec B = B\,\hat y,$$

i.e. $$\vec B$$ is along the $$+y$$-axis.

Step 5 – Ordinary language

The $$+y$$-axis corresponds to the vertically upward direction.

Result

The magnetic field in the chamber is vertically upward (from the floor towards the ceiling).

Concept involved : Ohm’s law relates the current $$I$$ flowing through a conductor to the potential difference $$V$$ applied across it and its resistance $$R$$ by the relation $$I = \frac{V}{R}$$. If the resistance of the path suddenly becomes extremely small, the current rises sharply.

Normal domestic circuit : In a lighting or power circuit the live wire carries the high potential, while the neutral wire is at zero potential. All appliances (lamp, fan, heater, etc.) are connected between these two conductors, so the current first passes through the appliance’s internal resistance before returning by the neutral wire.

How a short-circuit happens :

1. If the plastic insulation on the live and neutral conductors is damaged or melts, or if the two wires are joined accidentally, they may touch each other directly.
2. The current now bypasses the appliance and flows through the metallic conductors themselves, whose resistance is extremely small, practically $$R \approx 0\,\Omega$$.
3. Because $$R$$ is almost zero, Ohm’s law gives a huge current $$I = \frac{V}{R} \to \text{very large}$$.

The sudden flow of an excessively large current is called an electric short circuit. It may overheat the wiring and produce fire unless a fuse or an MCB disconnects the supply immediately.

Typical causes (for completeness)

• Frayed or cracked insulation.
• Loose connections allowing the wires to touch.
• Insertion of a conducting object (e.g. a nail) that bridges the two wires.

Thus, an electric short circuit occurs whenever the live and neutral (or positive and negative) conductors are connected directly together, providing a path of negligibly small resistance.

Step 1 ‒ Recall the three wires of a household circuit

• The live (or phase) wire is at a potential of about $$+220\,\text{V}$$ with respect to the ground.
• The neutral wire is at $$0\,\text{V}$$ because it is directly connected to the earth at the distribution transformer.
• The earth wire is an additional copper conductor that is also maintained at the earth potential $$0\,\text{V}$$, but it is connected only to the metallic body of an appliance and not to its working circuit.

Step 2 ‒ Function of the earth wire

The earth wire offers a path of very low resistance (typically a fraction of an ohm) between the metallic case and the ground. If, owing to wear of insulation or some defect, the live wire touches the metallic case, the potential of the case would instantly try to rise from $$0\,\text{V}$$ to $$+220\,\text{V}$$. Because of the earth wire, a large current $$I$$ now flows

$$I = \frac{V}{R_{\text{earth}}}$$

where $$V \approx 220\,\text{V}$$ and $$R_{\text{earth}}$$ is very small. This large current either

• immediately blows the fuse or trips the circuit-breaker, disconnecting the supply, and/or
• keeps the case at nearly $$0\,\text{V}$$ so a person touching it does not receive an electric shock.

Step 3 ‒ Why metallic appliances must be earthed

1. Most domestic gadgets (electric iron, refrigerator, heater, etc.) have an outer metallic body for mechanical strength and heat conduction.
2. If the insulation inside fails, the live conductor may come in contact with this body. Without earthing, the whole body would attain live potential; anyone touching it would then complete a circuit to the ground through his or her body and suffer a severe shock.
3. Earthing keeps the body at ground potential and ensures that the excess current chooses the path of least resistance—the earth wire—rather than the human body, which has a much higher resistance (of the order of $$10^3$$–$$10^5\,\Omega$$).

Final point

Thus, the earth wire is a vital safety feature: it prevents the metallic case from becoming live and causes the protective fuse or MCB to operate quickly, safeguarding both the user and the appliance.