Magnetometers

M/B_H from deflection, MB_H from oscillation.

12 minUpdated 22 October 2026deflection magnetometeroscillation magnetometerTan-A positionTan-B position

Lesson at a glanceShow the key points

Tan-A: M/B_H = (4π/μ₀)(d² − l²)² tan θ/2d (36.21); Tan-B: (4π/μ₀)(d² + l²)^(3/2) tan θ (36.22).
Null method compares moments; cot θ plots verify inverse square; B_H1/B_H2 = tan θ₂/tan θ₁.
Oscillation magnetometer: T = 2π√(I/MB_H) (36.23), MB_H = 4π²I/T² (36.24).
M = √(XY), B_H = √(Y/X) with X = M/B_H, Y = MB_H.

Two simple instruments, one measuring M/B_H and one measuring MB_H, together pin down both the strength of a magnet and the Earth’s horizontal field.

The deflection magnetometer

A small compass with a pointer over a 0°–90° scale (with a mirror against parallax), in a box mounted on a wooden frame with two long arms carrying metre scales that read distance from the needle’s centre. Its job is to measure M/B_H for a bar magnet, in one of two standard positions.

Tan-A position

Arms along magnetic east–west, pointer at 0°–0°. Lay the magnet along an arm, centre at distance d. The needle is end-on, so the magnet’s field (east or west) is perpendicular to B_H; the tangent law gives

Tan-A(36.21)

\frac{M}{B_H} = \frac{4\pi}{\mu_0}\,\frac{(d^2 - l^2)^2}{2d}\tan\theta

Tan-B position

Arms along magnetic north–south; the magnet lies across an arm, symmetric about it. Now the needle is broadside-on, the field again east–west:

Tan-B(36.22)

\frac{M}{B_H} = \frac{4\pi}{\mu_0}\,(d^2 + l^2)^{3/2}\tan\theta

Interactive

A deflection magnetometer

A compass at the centre of two long arms. Tan-A: arms east–west, magnet along the arm, so the compass sits end-on. Tan-B: arms north–south, magnet across the arm, so the compass sits broadside-on. Either way the magnet's field is perpendicular to B_H and the tangent law tan θ = B/B_H gives M/B_H.

deflection θ45.2°

field of the magnet35.3µT

M/B_H3.71e+4A m²/T

Position

Distance of the magnet

d

20.0cm

Magnetic moment

M

1.30A m²

Errors are removed by averaging 16 readings: both ends of the pointer (off-centre pivot), the magnet reversed end for end (magnetic centre ≠ geometric centre), turned upside down (magnetic axis ≠ geometric axis), and placed on the other arm (scale zero ≠ needle centre).

Worked example

M/B_H in Tan-A

A magnet of magnetic length 8.0 cm, its centre 20 cm from the needle in Tan-A position, gives a deflection of 45°. Find M/B_H.

Uses

Comparing two magnets by the null method: put one magnet on each arm and adjust distances until the needle doesn’t deflect; their fields at the needle are then equal. In Tan-B with short magnets, M₁/M₂ = d₁³/d₂³. No angle is measured, so it is both easier and more sensitive.
Testing the inverse-square law: in Tan-A, cot θ ∝ 2d/(d² − l²)², so plotting cot θ against (d² − l²)²/2d should give a straight line through the origin (in Tan-B, against (d² + l²)^(3/2)). Since these formulas come from B ∝ m/r² for a pole, the straight line verifies it.
Comparing B_H at two places: same magnet, same d: B_H1/B_H2 = tan θ₂/tan θ₁.

Worked example

Null method

Two short magnets on opposite arms in Tan-B position give no deflection when they are 20 cm and 25 cm from the needle. Compare their moments.

The oscillation magnetometer

A glass-walled box with a mirror on its base and a reference line along the middle; a hanger on an unspun silk thread from a torsion head. Level it, turn the box so the reference line lies in the magnetic meridian (check with a compass), take the twist out of the thread with a brass bar, then hang the magnet, north end north, and set it swinging gently.

Displaced by θ, the magnet feels a restoring torque −MB_H sin θ ≈ −MB_Hθ (ignoring the thread’s twist). With moment of inertia I, that is angular SHM:

I\alpha = -MB_H\theta \quad\Rightarrow\quad \omega = \sqrt{\frac{MB_H}{I}}

Oscillation magnetometer(36.23–36.24)

T = 2\pi\sqrt{\frac{I}{MB_H}}, \qquad MB_H = \frac{4\pi^2 I}{T^2}

for a rectangular bar of mass W, length a, breadth b: I = W(a² + b²)/12

Interactive

An oscillation magnetometer

A magnet hung by an unspun silk thread swings about the north–south line. For small angles the restoring torque is −MB_Hθ, so it oscillates in SHM with T = 2π√(I/MB_H). Measuring T gives MB_H; combined with M/B_H from a deflection magnetometer, both M and B_H follow.

T = 2\pi\sqrt{\frac{I}{MB_H}}

5.02s

MB_H = \frac{4\pi^2I}{T^2}

4.55×10⁻⁵ N m

Moment of inertia

I

2.9×10⁻⁵ kg m²

Magnetic moment

M

1.30A m²

Horizontal field

B_H

3.5×10⁻⁵ T

Worked example

Comparing the Earth's field

A compass needle makes 20 oscillations a minute where the dip is 30°, and 24 a minute where the dip is 50°. Compare the total field at the two places.

Finding M and the horizontal field separately

With X = M/B_H from a deflection magnetometer and Y = MB_H from an oscillation magnetometer:

Magnet moment and Earth's field

M = \sqrt{XY}, \qquad B_H = \sqrt{Y/X}

Worked example

Putting them together

The magnet above (M/B_H = 3.69 × 10⁴ A m²/T) has I = 2.9 × 10⁻⁵ kg m² and oscillates with T = 5.0 s. Find M and B_H.

What to remember

1Deflection magnetometer: M/B_H from Tan-A (36.21) or Tan-B (36.22) with the tangent law.
2Null method compares moments; cot θ plots verify the inverse-square law; tan θ ratios compare B_H.
3Oscillation magnetometer: T = 2π√(I/MB_H) (36.23).
4M = √(XY), B_H = √(Y/X).

Check yourself

A magnet oscillates with period T. At a place where B_H is four times larger, the period becomes…

Why is the null method better for comparing two magnets?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadMagnetometerWikipedia

PreviousGalvanometers Next Gauss's Law for Magnetism

Magnetometers

M/B_H from deflection, MB_H from oscillation.

12 minUpdated 22 October 2026deflection magnetometeroscillation magnetometerTan-A positionTan-B position

Lesson at a glanceShow the key points

Tan-A: M/B_H = (4π/μ₀)(d² − l²)² tan θ/2d (36.21); Tan-B: (4π/μ₀)(d² + l²)^(3/2) tan θ (36.22).
Null method compares moments; cot θ plots verify inverse square; B_H1/B_H2 = tan θ₂/tan θ₁.
Oscillation magnetometer: T = 2π√(I/MB_H) (36.23), MB_H = 4π²I/T² (36.24).
M = √(XY), B_H = √(Y/X) with X = M/B_H, Y = MB_H.

Two simple instruments, one measuring M/B_H and one measuring MB_H, together pin down both the strength of a magnet and the Earth’s horizontal field.

The deflection magnetometer

Tan-A position

Tan-A(36.21)

\frac{M}{B_H} = \frac{4\pi}{\mu_0}\,\frac{(d^2 - l^2)^2}{2d}\tan\theta

Tan-B position

Arms along magnetic north–south; the magnet lies across an arm, symmetric about it. Now the needle is broadside-on, the field again east–west:

Tan-B(36.22)

\frac{M}{B_H} = \frac{4\pi}{\mu_0}\,(d^2 + l^2)^{3/2}\tan\theta

Interactive

A deflection magnetometer

deflection θ45.2°

field of the magnet35.3µT

M/B_H3.71e+4A m²/T

Position

Distance of the magnet

d

20.0cm

Magnetic moment

M

1.30A m²

Worked example

M/B_H in Tan-A

A magnet of magnetic length 8.0 cm, its centre 20 cm from the needle in Tan-A position, gives a deflection of 45°. Find M/B_H.

Uses

Comparing two magnets by the null method: put one magnet on each arm and adjust distances until the needle doesn’t deflect; their fields at the needle are then equal. In Tan-B with short magnets, M₁/M₂ = d₁³/d₂³. No angle is measured, so it is both easier and more sensitive.
Testing the inverse-square law: in Tan-A, cot θ ∝ 2d/(d² − l²)², so plotting cot θ against (d² − l²)²/2d should give a straight line through the origin (in Tan-B, against (d² + l²)^(3/2)). Since these formulas come from B ∝ m/r² for a pole, the straight line verifies it.
Comparing B_H at two places: same magnet, same d: B_H1/B_H2 = tan θ₂/tan θ₁.

Worked example

Null method

Two short magnets on opposite arms in Tan-B position give no deflection when they are 20 cm and 25 cm from the needle. Compare their moments.

The oscillation magnetometer

Displaced by θ, the magnet feels a restoring torque −MB_H sin θ ≈ −MB_Hθ (ignoring the thread’s twist). With moment of inertia I, that is angular SHM:

I\alpha = -MB_H\theta \quad\Rightarrow\quad \omega = \sqrt{\frac{MB_H}{I}}

Oscillation magnetometer(36.23–36.24)

T = 2\pi\sqrt{\frac{I}{MB_H}}, \qquad MB_H = \frac{4\pi^2 I}{T^2}

for a rectangular bar of mass W, length a, breadth b: I = W(a² + b²)/12

Interactive

An oscillation magnetometer

T = 2\pi\sqrt{\frac{I}{MB_H}}

5.02s

MB_H = \frac{4\pi^2I}{T^2}

4.55×10⁻⁵ N m

Moment of inertia

I

2.9×10⁻⁵ kg m²

Magnetic moment

M

1.30A m²

Horizontal field

B_H

3.5×10⁻⁵ T

Worked example

Comparing the Earth's field

A compass needle makes 20 oscillations a minute where the dip is 30°, and 24 a minute where the dip is 50°. Compare the total field at the two places.

Finding M and the horizontal field separately

With X = M/B_H from a deflection magnetometer and Y = MB_H from an oscillation magnetometer:

Magnet moment and Earth's field

M = \sqrt{XY}, \qquad B_H = \sqrt{Y/X}

Worked example

Putting them together

The magnet above (M/B_H = 3.69 × 10⁴ A m²/T) has I = 2.9 × 10⁻⁵ kg m² and oscillates with T = 5.0 s. Find M and B_H.

What to remember

1Deflection magnetometer: M/B_H from Tan-A (36.21) or Tan-B (36.22) with the tangent law.
2Null method compares moments; cot θ plots verify the inverse-square law; tan θ ratios compare B_H.
3Oscillation magnetometer: T = 2π√(I/MB_H) (36.23).
4M = √(XY), B_H = √(Y/X).

Check yourself

A magnet oscillates with period T. At a place where B_H is four times larger, the period becomes…

Why is the null method better for comparing two magnets?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadMagnetometerWikipedia

PreviousGalvanometers Next Gauss's Law for Magnetism

The deflection magnetometer#

Tan-A position

Tan-B position

Uses

The oscillation magnetometer#

Finding M and the horizontal field separately#

Check yourself

Keep exploring

The deflection magnetometer#

Tan-A position

Tan-B position

Uses

The oscillation magnetometer#

Finding M and the horizontal field separately#

Check yourself

Keep exploring

The deflection magnetometer

The oscillation magnetometer

Finding M and the horizontal field separately

The deflection magnetometer

The oscillation magnetometer

Finding M and the horizontal field separately