Q.The magnetic field at the centre O of the following current loop is: the loop consists of two straight radial segments meeting a semicircular arc of radius r centred at O -- current I enters along one straight radial lead, traces the semicircular arc, and leaves along the other straight radial lead (both leads point directly at O, and a current element aimed directly at the field point contributes zero field there by the Biot-Savart law).
You already know that a stationary charge creates an electric field that falls off as 1/r2 and points radially away from the charge. But when that charge moves — when it becomes a current — something new appears: a magnetic field. The question is: how does a tiny piece of current produce a tiny piece of magnetic field?
Imagine a very short segment of wire carrying a steady current I. Let its length be dl — so small that we can treat it as a point-like source. This little current element, Idl, is the magnetic analogue of a point charge in electrostatics. Just as Coulomb’s law gives the electric field from a point charge, the Biot-Savart law gives the magnetic field from a current element.
But there’s a crucial difference. The electric field from a point charge points along the line joining the charge to the observation point. The magnetic field from a current element points perpendicular to both the direction of the current and the line joining the element to the point. This perpendicular nature is the heart of magnetism.
The Precise Statement
Consider a current element Idl located at some point. Let r be the position vector from the element to the point P where we want the magnetic field. Then the infinitesimal magnetic field dB at P due to this element is:
dB=4πμ0r2Idl×r^
Here:
μ0=4π×10−7T⋅m/A is the permeability of free space — a fundamental constant.
dl points along the direction of the current.
r^ is a unit vector pointing from the current element to the observation point.
The cross product dl×r^ gives both the magnitude and direction.
What the Cross Product Tells You
The magnitude of the cross product is ∣dl×r^∣=dl⋅1⋅sinθ, where θ is the angle between dl and r^. So the magnitude of dB is:
dB=4πμ0r2Idlsinθ
This is exactly the form you mentioned: proportional to Idlsinθ/r2. The sinθ factor means:
When the current element points directly toward or away from P (θ=0 or π), sinθ=0 — no magnetic field is produced along that line.
When the current element is perpendicular to the line joining it to P (θ=90∘), the field is maximum.
The direction of dB is given by the right-hand rule: curl the fingers of your right hand from dl toward r^, and your thumb points in the direction of dB. This direction is always perpendicular to the plane containing dl and r.
Watch out
A common mistake is to think dB points along r or along dl. It does neither — it is perpendicular to both. If you ever find yourself drawing dB in the plane of the page when dl and r are also in the page, you are wrong: dB comes out of or goes into the page.
Why the 1/r2 Dependence?
Just like Coulomb’s law, the Biot-Savart law has an inverse-square dependence on distance. This is not a coincidence — both laws emerge from the same underlying structure of electromagnetism. Unlike Coulomb's law, this 1/4π prefactor is not because the field spreads uniformly over a sphere -- the sinθ factor above already shows the elemental field is NOT isotropic, it circulates around the current direction instead. The 1/(4π) here is simply a consequence of the SI 'rationalized' unit convention, chosen so that μ0 appears without a 4π in Ampere's circuital law, ∮B⋅dl=μ0Ienc.
The Total Field: Integration
The Biot-Savart law gives you the field from a single infinitesimal current element. To find the total magnetic field from a complete circuit (a wire of any shape), you must integrate over the entire path:
B=4πμ0∫r2Idl×r^
This integral is a vector sum — you add up the contributions from every tiny segment, each with its own direction. This is why the Biot-Savart law is powerful: it lets you compute the magnetic field of any current-carrying wire, from a straight wire to a circular loop to a solenoid.
Tip
For a straight infinite wire, the integration yields B=2πrμ0I, where r is the perpendicular distance from the wire. For a circular loop of radius R at its centre, B=2Rμ0I. These are standard results you should remember — but always derive them from the Biot-Savart law at least once.
The Big Picture
The Biot-Savart law is to magnetism what Coulomb’s law is to electrostatics. It tells you how a moving charge (a current) creates a magnetic field. The field is always perpendicular to both the current direction and the line joining the source to the point of observation. This perpendicular nature is why magnetic fields can do things electric fields cannot — like exert forces on moving charges in directions perpendicular to their motion, leading to circular paths and cyclotron motion.
The Biot-Savart law: dB=4πμ0r2Idl×r^ — the fundamental rule for how currents create magnetic fields.
The Biot-Savart law is one of the most important derivation-based topics in the CBSE Class 12 Physics NCERT curriculum under Moving Charges and Magnetism, frequently searched as Biot-Savart law derivation and formula class 12 or Biot-Savart law important questions. Since it is the starting point for nearly every magnetic-field formula tested in JEE Main and NEET physics, this concept is genuinely foundational, not just a board-exam checkbox.
The two straight radial leads point directly at the centre O, so by the Biot-Savart law they contribute nothing there; only the semicircular arc contributes, giving half of a full loop's field.
✓Final answer
(a) 4rμ0I, into the page
Step 1. Split the current path into its three pieces: the straight lead in, the semicircular arc of radius r, and the straight lead out.
Step 2. For each straight lead, every current element Idl points directly toward (or away from) the centre O, so the angle θ between dl and r^ is 0° or 180°, and sinθ=0. By the Biot-Savart law, dB=4πμ0r2Idlsinθ=0 for every element of both straight leads -- they contribute nothing to the field at O.
Step 3. The full field at O therefore comes only from the semicircular arc. For a complete circular loop of radius r, the centre field is Bloop=μ0I/2r (§3.8.3). A semicircle is exactly half of a full loop, and every element of the semicircle is at the same distance r from O with θ=90°, so its contribution is exactly half of the full loop's:
B=21×2rμ0I=4rμ0I
Step 4. Direction: applying the right-hand thumb rule to the sense of current flow around the arc gives a field into the page at O (matching option (a)).
✓Final answer
(a) 4rμ0I, into the page
Straight leads pointing at the centre contribute zero field there; only the semicircular arc contributes, giving half the full-loop field.
Forgetting that a current element aimed directly at the field point gives sin(theta)=0 and hence zero field.
Using the full-loop formula mu0 I/2r instead of halving it for a semicircular arc.