Q.A long straight wire carries a current of $15\ \text{A}$. Find the magnetic field at a point $20\ \text{cm}$ from the wire.

The field of a straight wire at a point is along $\hat{I} \times \hat{r}$. With the current along $-\hat{j}$ and the point on the positive $x$-axis ($\hat{r} = +\hat{i}$), this gives $(-\hat{j}) \times \hat{i} = +\hat{k}$ — the field points along the $z$-axis, option (A).

By the right-hand (thumb) rule, the magnetic field produced by a current flowing from South to North, at a point directly below the wire, points towards the West — so the compass needle (normally pointing North) deflects towards the West, i.e. to the left, when viewed in the conventional map orientation (North at top, East to the right).

Setting up directions

Take the standard compass/map convention: North $=+\hat y$, East $=+\hat x$, Up (vertically, away from the ground) $=+\hat z$. This is a right-handed set: $\hat x\times\hat y=\hat z$.

The wire is aligned along the North–South line and carries current from S to N, i.e. the current direction is $\hat I = +\hat y$ (Northward).

The compass needle sits directly below the wire — i.e. at a point displaced from the wire in the $-\hat z$ direction (vertically downward, towards the ground).

Applying the Biot–Savart / right-hand rule

For an infinite straight current-carrying wire, the magnetic field at a point is directed along $\hat I \times \hat r$, where $\hat r$ is the unit vector from the wire to the field point.

Here $\hat r = -\hat z$ (point is directly below the wire), so:

$\vec B \ \propto\ \hat I\times\hat r = \hat y\times(-\hat z) = -(\hat y\times\hat z) = -\hat x$

Since $+\hat x=$ East, $-\hat x=$ West. So the magnetic field produced by the wire, at the point directly below it, points towards the West.

Effect on the compass needle

Before the current flows, the compass needle's north pole points geographic North (Earth's field). When the current flows from S to N, it adds a magnetic field pointing West at the needle's location. The needle aligns with the resultant of "Earth's field (North)" and "wire's field (West)" — so its north pole swings away from due North, towards the West.

In the conventional way such a scene is drawn/viewed (looking down at the ground from above, with North at the top of the page and East to the right, exactly matching the given figure where N is at the top and current flows upward), West corresponds to the left-hand side of the page. So the needle deflects to the left.

(This is a modern restatement of Oersted's original 1820 observation, which first revealed the link between electric current and magnetism.)

At the midpoint between two parallel wires carrying equal currents in the same direction, each wire's field has equal magnitude but the right-hand rule shows the two fields point in opposite directions there — so they cancel and the net field is zero.

Setup

Two long, straight, parallel wires, separated by $2r$, each carrying the same current $I$ in the same direction. Consider the point exactly midway between them — a distance $r$ from each wire.

Field due to each wire

For an infinite straight wire, the field at perpendicular distance $r$ has magnitude

$B=\frac{\mu_0 I}{2\pi r}$

directed in circles around the wire (by the right-hand rule — point the thumb along the current, fingers curl in the direction of $\vec B$).

So each wire, individually, produces a field of the same magnitude $\dfrac{\mu_0 I}{2\pi r}$ at the midpoint (since both are the same distance $r$ from it and carry the same current $I$).

Direction of each field at the midpoint

Let the two wires be parallel to each other, both carrying current in (say) the $+\hat z$ direction, with wire 1 at $x=0$ and wire 2 at $x=2r$; the midpoint is at $x=r$.

• For wire 1, the midpoint lies in the $+\hat x$ direction from the wire. By the right-hand rule ($\vec B \propto \hat I\times\hat r$), $\hat z\times\hat x=\hat y$, so $\vec B_1$ points in $+\hat y$.
• For wire 2, the midpoint lies in the $-\hat x$ direction from that wire. So $\hat z\times(-\hat x)=-\hat y$, i.e. $\vec B_2$ points in $-\hat y$.

So $\vec B_1$ and $\vec B_2$ are equal in magnitude but opposite in direction at the midpoint.

Net field

$\vec B_{\text{net}} = \vec B_1+\vec B_2 = \frac{\mu_0 I}{2\pi r}\hat y - \frac{\mu_0 I}{2\pi r}\hat y = 0$

(This is the key difference from the case of opposite currents, where the two fields at the midpoint would instead add up.)

The magnetic field around a long straight current-carrying wire forms concentric circles in a plane perpendicular to the wire. For a vertical wire with upward current, the field lines are horizontal circles centered on the wire — so the correct option is (C).

(b) (i) – (r), (ii) – (q), (iii) – (p). Straight wire $\to \dfrac{\mu_0 I}{2\pi r}$ (r); circular-loop centre $\to \dfrac{\mu_0 I}{2r}$ (q); solenoid $\to \mu_0 n I$ (p).

Deriving μ from the force law between two current-carrying wires gives the dimensional formula [M L T⁻² A⁻²].

The force per unit length between two long parallel current-carrying wires (carrying currents I_1, I_2, separated by distance d) is:

F/L = μ I_1 I_2 / (2π d)

So μ = 2π d (F/L) / (I_1 I_2). We find the dimensions of each quantity on the right:

• F (force) has dimensions [M L T⁻²]
• d and L (lengths) have dimensions [L]
• I_1, I_2 (currents) have dimensions [A]

Substituting:

[μ] = [L] × [M L T⁻²] / [L] ÷ [A²]

= [M L T⁻²] / [A²]

= [M L T⁻² A⁻²]

This matches the known SI value of the permeability of free space, μ_0 = 4π×10⁻⁷ T·m/A = 4π×10⁻⁷ kg·m·s⁻²·A⁻².

Hans Christian Oersted discovered the magnetic effect of current.

In 1820 Oersted noticed that a compass needle near a current-carrying wire deflected, showing that an electric current produces a magnetic field. This observation, foundational to the NCERT/CBSE Class 12 moving charges and magnetism chapter, later led to the Biot–Savart and Ampere laws.

At a point equidistant from two parallel current-carrying wires, the individual fields either cancel or add depending on the current directions and O's position relative to the wires.

For two long straight parallel wires each carrying the same current I, each produces a field of magnitude B at point O (equidistant from both wires).

• When the geometry is such that the two individual fields at O point in opposite directions (as happens, for example, when O lies symmetrically between two wires carrying current in the same direction), they cancel: resultant = zero.
• When the geometry is such that the two individual fields at O point in the same direction (as happens, for example, when the currents are in opposite directions, or O is positioned differently relative to the wires), they add: resultant = 2B.

Note on the figure: the scanned description could not fully resolve the exact wire spacing/position of O in each case, so the precise Case-I/Case-II geometry is taken from the standard, widely-used version of this question, matching the answer below.

Using the right-hand rule for a current flowing west, the field directly above the wire points north.

Take east = $\hat{x}$, north = $\hat{y}$, up = $\hat{z}$ (a right-handed set, since east × north = up). The current flows east to west, i.e. along $-\hat{x}$.

Grip the wire with the right hand, thumb pointing west (direction of current). The fingers curl to show the field direction. At a point directly above the wire, this curl points towards north.

Vector check: field direction $\propto \hat{I}\times\hat{r}$, where $\hat{r}$ points from the wire to the field point. Here $\hat{I}=-\hat{x}$ and $\hat{r}=\hat{z}$ (point is above). $(-\hat{x})\times\hat{z} = -(\hat{x}\times\hat{z}) = -(-\hat{y}) = \hat{y}$, i.e. north.

The magnetic field lines of an infinite straight current-carrying wire are concentric circles around the wire, in a plane perpendicular to it.

For an infinitely long, straight conductor carrying current $I$, the magnetic field at a perpendicular distance $r$ from the wire has magnitude:

$B = \frac{\mu_0 I}{2\pi r}$

By the right-hand (or Maxwell's corkscrew) rule, if the thumb points along the direction of current flow, the curled fingers show the direction of the magnetic field. Since $B$ has the same magnitude at every point on a circle of radius $r$ centred on the wire, and is always tangential to that circle, the field lines trace out concentric circles in planes perpendicular to the wire, with the wire passing through their common centre.

Two standard rules used to find the direction of the magnetic field produced by a current: the Right-Hand Thumb Rule and Maxwell's Cork-Screw Rule.

1. Right-Hand Thumb Rule (Right-Hand Grip Rule): If a current-carrying conductor is held in the right hand such that the thumb points in the direction of current flow, the direction in which the fingers curl around the conductor gives the direction of the magnetic field lines around it.

2. Maxwell's Cork-Screw Rule: If a right-handed screw is imagined to advance in the direction of the current, the direction in which the screw's head must be rotated gives the direction of the magnetic field lines around the conductor.

(Both rules give identical results and are used interchangeably for straight wires, loops, and solenoids.)

By Ampere's circuital law, the field of an infinite straight current-carrying wire is $B=\mu_0I/(2\pi R)$, which is inversely proportional to the distance $R$.

Working

Applying Ampere's circuital law to a circular Amperian loop of radius $R$ centred on a long straight wire carrying current $I$ (by symmetry, $B$ is uniform in magnitude around the loop and tangential to it):

$\oint \vec B\cdot d\vec l=\mu_0I\ \Rightarrow\ B(2\pi R)=\mu_0I\ \Rightarrow\ B=\dfrac{\mu_0I}{2\pi R}$

Since $\mu_0$ and $I$ are constant, $B\propto\dfrac1R$.