This MCQ module is based on: Biot Savart Amperes Law
Biot Savart Amperes Law
This assessment will be based on: Biot Savart Amperes Law
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Biot Savart Amperes Law
4.5 Magnetic Field due to a Current Element - Biot-Savart Law
Just as Coulomb's law gives the electric field of a point charge, the Biot-Savart law gives the elementary magnetic field produced by an infinitesimal current element \(I\,d\vec L\) at a point P located at a distance r:
where \(\mu_0 = 4\pi\times10^{-7}\) T m A\(^{-1}\) is the permeability of free space and \(\hat r\) is the unit vector from the element to P.
Magnitude: \(dB = \dfrac{\mu_0}{4\pi}\,\dfrac{I\,dL\,\sin\theta}{r^2}\). Direction: along \(d\vec L \times \hat r\), found by the right-hand rule.
- Both fall off as \(1/r^2\).
- Both linear in source strength (q for E, I dL for B).
- E is along \(\hat r\); B is perpendicular to both \(d\vec L\) and \(\hat r\) (cross product).
- \(\mu_0/4\pi = 10^{-7}\) T m/A vs \(1/(4\pi\varepsilon_0) = 9\times10^9\) N m\(^2\)/C\(^2\).
| Aspect | Biot-Savart Law | Ampere's Law |
|---|---|---|
| Form | dB = (mu0/4 pi) I dL x r-hat / r² | integral B . dL = mu0 I_enc |
| Type | Differential (point source) | Integral (closed loop) |
| Best for | Any geometry, especially finite wires & loops | Highly symmetric problems |
| Analogue | Coulomb's law | Gauss's law |
| Universal validity | Yes | Yes (with displacement-current correction) |
4.6 Magnetic Field on the Axis of a Circular Current Loop
Consider a circular loop of radius R carrying current I, with its centre at the origin and axis along x. Apply Biot-Savart to a small element on the loop at distance \(r = \sqrt{R^2+x^2}\) from a point P on the axis. Symmetry kills the components perpendicular to the axis; only the axial components add.
Special case - centre of the loop (x = 0):
For N closely wound turns, multiply by N: \(B = \mu_0 NI/(2R)\) at the centre.
Worked Example 4.4 - Field at the centre of a coil
A circular coil of 200 turns and radius 10 cm carries 0.4 A. Find the magnetic field at its centre.
\(B = \dfrac{\mu_0 NI}{2R} = \dfrac{(4\pi\times10^{-7})(200)(0.4)}{2(0.10)}\)
= \(5.03\times10^{-4}\) T \(\approx 5.0\) gauss.
4.7 Ampere's Circuital Law
For static currents, Ampere proved an extremely useful integral relation - the magnetic analogue of Gauss's law:
The line integral of \(\vec B\) around any closed loop equals \(\mu_0\) times the total current threading through any surface bounded by that loop. The closed loop is called an Amperian loop.
4.7.1 Field of a Long Straight Wire
By symmetry B has the same magnitude on a circle of radius r around the wire and is tangent to it. Therefore
Worked Example 4.5 - Field of a power line
A long straight wire carries 35 A. What is the magnitude of B at 20 cm from the wire?
\(B = \dfrac{\mu_0 I}{2\pi r} = \dfrac{(4\pi\times10^{-7})(35)}{2\pi(0.20)} = 3.5\times10^{-5}\) T = 35 microtesla.
About the same as Earth's field - a real concern for sensitive instruments near power lines.
4.8 The Solenoid
A solenoid is a long helical coil. If turns per unit length is n and current is I, applying Ampere's law to a rectangular loop with one side inside and one side far outside gives:
4.8.1 The Toroid
A toroid is a solenoid bent into a closed circle. Apply Ampere's law to a circular Amperian loop of radius r inside the windings (N total turns):
Worked Example 4.6 - Solenoid field
A solenoid 0.6 m long has 600 turns and carries 3.0 A. Find B inside.
n = 600/0.6 = 1000 turns/m.
B = mu0 n I = (4 pi x 10-7)(1000)(3.0) = 3.77 x 10-3 T ~ 38 gauss.
Interactive: Magnetic Field along the Axis of a Loop L4 Analyse
Slide the field point along the axis. Watch B fall away from the centre as B = mu0 I R2/(2(R2+x2)3/2).
Place a compass under a long wire connected to a battery (use a series resistor for safety). Switch on the current.
The compass underneath deflects to the east. By the right-hand rule, with thumb pointing N (current direction), fingers curl down on the south side of the wire and up on the north side; below the wire B points east, deflecting the compass needle from north toward east.
Reverse the battery: needle deflects west. This is exactly Oersted's 1820 discovery.
Competency-Based Questions L3-L5
Q1. Which law most efficiently gives the field inside the long solenoid?
Q2. (Numerical) What current is needed in the MRI solenoid to produce 1.5 T?
Q3. (Fill in the blank) The magnetic field a perpendicular distance r from a long straight wire is B = ___.
Q4. (True/False) For a toroid, the magnetic field outside the core is essentially zero.
Q5. (HOT) Two identical loops are placed coaxially distance R apart (one of radius R). Show that the field at the midpoint is more uniform than at the centre of a single loop. (Helmholtz coils)
Assertion-Reason Questions L4 Analyse
(a) Both A and R true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
A: Inside an ideal long solenoid the magnetic field is uniform.
R: Each turn behaves like a small magnetic dipole and the dipole fields add constructively along the axis.
A: Ampere's law applies to any closed loop, regardless of geometry.
R: Ampere's law gives a useful direct result only when the field has high symmetry.
A: A circular current loop behaves like a magnetic dipole.
R: At large distances on the axis, B falls as 1/x³, characteristic of a dipole field.
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Board exam sample papers
Physics — CBSE Class XII Sample Paper 1 (2025-26)
Section A · Section B · Section C · Section D · Section E