Q.If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30∘ with the direction of applied field?
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Magnetic Poles: The Intuition First
Imagine you have a bar magnet — the kind you might have stuck on your refrigerator. If you bring two of them close, something interesting happens. Sometimes they snap together with a satisfying click. Other times, they push each other away, refusing to touch no matter how hard you try.
That's not random. Every magnet has two special regions, one at each end, where the magnetic force is strongest. These are its magnetic poles.
The word "pole" comes from the Greek polos, meaning "pivot" or "axis" — the Earth itself has a North Pole and a South Pole, and it behaves like a giant magnet.
The Two Types of Poles
Every magnet has exactly two poles: a north pole and a south pole. You cannot have a magnet with only one pole — cut a bar magnet in half, and each half immediately becomes a complete magnet with its own north and south poles.
The rule of interaction is simple and memorable:
- Unlike poles attract: north pulls south, south pulls north.
- Like poles repel: north pushes north away; south pushes south away.
This is the fundamental behaviour. No exceptions.
The Precise Statement
Magnetic poles are the regions of a magnet where the external magnetic field is strongest. Every magnet has exactly two poles — a north pole and a south pole — that cannot be isolated. Like poles repel; unlike poles attract.
The key points to remember for exams:
- Poles always come in pairs — there is no magnetic monopole (a single isolated pole) in nature, despite decades of searching.
- The north pole is defined as the pole that points toward Earth's geographic north when the magnet is freely suspended.
- The south pole points toward Earth's geographic south.
A Common Confusion (Watch Out)
Earth's geographic North Pole is actually a magnetic south pole. Why? Because the north pole of a compass needle (which is a magnetic north pole) is attracted to it. And unlike poles attract. So the Earth's north pole behaves like a magnetic south pole. This often trips students up in exams.
Why This Matters …
Why this formula?
Magnetic Poles: Why the Key Formulas Hold
Let's build this from first principles — understanding why a magnetic pole behaves the way it does, not just memorizing the result.
1. What Is a Magnetic Pole?
A magnetic pole is a conceptual point where the magnetic field appears to originate or terminate. In reality, magnetic poles always come in north-south pairs (no isolated monopoles exist in nature), but we treat them as idealized sources for calculations.
- North pole: source of magnetic field lines (outward)
- South pole: sink of magnetic field lines (inward)
2. The Key Formula: Force Between Two Magnetic Poles
The force between two magnetic poles of strengths m1 and m2, separated by distance r, is:
F=4πμ0⋅r2m1m2
Why this form?
This is a Coulomb's law analog — and that's not a coincidence. Here's the reasoning:
-
Experimental observation: Magnetic poles attract/repel with a force that:
- Varies as 1/r2 (inverse square law)
- Is proportional to the product of pole strengths
- Depends on the medium (via μ0, the permeability of free space)
-
Mathematical analogy: The magnetic field B at distance r from a single pole m is:
B=4πμ0⋅r2m
This comes from Gauss's law for magnetism applied to a point source.
- Force derivation: The force on pole m2 in the field of pole m1 is:
F=m2⋅B1=m2⋅(4πμ0⋅r2m1)
Hence:
F=4πμ0⋅r2m1m2
Key insight: The 1/r2 dependence is not arbitrary — it follows from the geometry of 3D space (flux spreads over a sphere of area 4πr2).
3. The Magnetic Field of a Bar Magnet (Two Poles)
For a bar magnet of length 2l with poles +m and −m, the field at a point on the axis at distance x from the center is:
B=4πμ0⋅(x2−l2)22ml
Why this form?
-
Superposition principle: The total field is the vector sum of fields from the north pole (+m) and south pole (−m).
-
Field from north pole at distance (x−l):
BN=4πμ0⋅(x−l)2m(away from north)
- Field from south pole at distance (x+l):
BS=4πμ0⋅(x+l)2m(toward south)
- Net field (both along same direction on axis):
B=BN−BS=4πμ0m[(x−l)21−(x+l)21]
- Simplify using algebra:
(x−l)21−(x+l)21=(x2−l2)24xl
Therefore:
B=4πμ0⋅(x2−l2)24mxl
But for a bar magnet, the magnetic moment is M=m⋅(2l) (pole strength × separation). So 2ml=M, giving:
B=4πμ0⋅(x2−l2)22Mx
Key insight: The field is not simply 1/r2 because we have two poles — the net effect is a dipole field, which falls off as 1/r3 at large distances.
4. The Far-Field Approximation (Dipole Formula)
For x≫l (far from the magnet), x2−l2≈x2, so:
B≈4πμ0⋅x32M
Why 1/x3?
- A single pole gives 1/r2 …
Concept: Magnetic Poles — a current-carrying solenoid behaves like a bar magnet with a magnetic moment m.
Step 1: From Exercise 5.5, the magnetic moment of the solenoid is
m=0.6 J/T.
Step 2: Torque on a magnetic dipole in a uniform field is
τ=mBsinθ, where θ is the angle between m and B.
Step 3: Here B=0.25 T, θ=30∘, so …
The solenoid of Exercise 5.5 (800 turns, area 2.5×10−4 m2, current 3.0 A) has magnetic moment m=NIA=0.60 J T−1. In a field B=0.25 T at θ=30∘, the torque is τ=mBsinθ=7.5×10−2 Nm.
Step-by-Step Solution
Magnetic moment of the solenoid (from the Exercise 5.5 data N=800, I=3.0 A, A=2.5×10−4 m2):
m=NIA=800×3.0×2.5×10−4=0.60 J T−1.
Torque on a magnetic moment in a uniform field:
τ=mBsinθ, …
Method: Torque on a Magnetic Dipole in a Uniform Magnetic Field
This problem uses the magnetic dipole torque formula — the solenoid behaves like a bar magnet with a magnetic moment.
Step-by-step solution
Step 1: Recall the torque formula
For a magnetic dipole (or solenoid) in a uniform magnetic field:
τ=M×B
Magnitude:
τ=MBsinθ
where:
- M = magnetic moment of the solenoid
- B = applied magnetic field strength
- θ = angle between the solenoid axis and the field direction
Step 2: Identify given values
From Exercise 5.5 (assumed context), the magnetic moment of the solenoid is:
M=0.6 A m2
Given in this problem:
- B=0.25 T
- θ=30∘
Step 3: Apply the formula
τ=(0.6)(0.25)sin30∘
Since sin30∘=21: …
Here are the common mistakes students make on this magnetic torque problem, along with how to avoid each.
1. Using the Wrong Formula for Torque
Mistake:
Students often apply τ=mB (the formula for force on a current-carrying wire) or τ=NIAB without the sinθ factor.
Why it’s wrong:
Torque on a magnetic dipole (like a solenoid) in a uniform field is given by:
τ=mBsinθ
where m=NIA is the magnetic moment of the solenoid.
How to avoid:
Always write the full vector form first: τ=m×B. Then extract magnitude as τ=mBsinθ. Never drop the sinθ term.
2. Confusing the Angle θ
Mistake:
Using θ=30∘ directly in sinθ when the problem says the axis makes 30∘ with the field.
Why it’s correct here:
The angle between the magnetic moment vector (along the solenoid axis) and the field is exactly 30∘. So θ=30∘ is correct.
How to avoid:
Draw a diagram. Mark the solenoid axis, the field direction, and the angle between them. That angle is θ in τ=mBsinθ.
3. Forgetting to Calculate m=NIA First
Mistake:
Plugging numbers directly into τ=NIABsinθ without computing m — or using wrong values for N, I, or A.
Why it’s wrong:
The torque depends on the product NIA, not just I or A alone.
How to avoid:
List given data:
- N = number of turns
- I = current
- A = cross-sectional area (πr2 for circular)
Then compute m=NIA separately before plugging into torque formula.
4. Unit Errors (Area, Field, Torque)
Mistake:
Using area in cm2 without converting to m2, or mixing Tesla with Gauss.
How to avoid:
Always convert to SI units:
- 1 cm2=10−4 m2
- 1 G=10−4 T …
- CBSE 2026Set ANNUAL1 markQ.If magnetic monopoles existed then write the equation of Gauss's law for magnetism.
›Reveal solutionSolution
Currently Gauss's law for magnetism states the net magnetic flux through any closed surface is zero (no monopoles); if monopoles existed, the right side would instead equal mu0 times the enclosed pole strength, just like the electric case.
Gauss's law for magnetism in its present form is: the closed surface integral (over any closed surface S) of B . dA = 0, reflecting the experimental fact that isolated magnetic poles (monopoles) have never been observed - magnetic field lines always form closed loops with no starting/ending point (no magnetic 'charge'). If magnetic monopoles did exist, with an isolated pole strength qm enclosed by the surface, this law would take a form exactly analogous to Gauss's law for electric charge: …
- CBSE 2025Set ANNUAL1 markMCQQ.The value of magnetic induction at a distance r from a single pole is inversely proportional to(a) r(b) r^2(c) 1/r(d) 1/r^2
›Reveal solutionSolution
An isolated magnetic pole of strength m produces B = (mu0/4pi)(m/r^2), an inverse-square law, so the magnetic induction is inversely proportional to r^2.
The magnetic field (magnetic induction) at a distance r from a single magnetic pole of pole strength m is given by Coulomb's law of magnetism:
B = (mu0 / 4*pi) * (m / r^2)
…
- CBSE 2025Set ANNUAL1 markMCQQ.The ultimate individual unit of magnetism in any magnet is :(a) north pole(b) south pole(c) magnetic dipole(d) quadrupole
›Reveal solutionSolution
Isolated magnetic monopoles (single north or south poles) do not exist in nature; magnetic poles always occur in pairs, so the magnetic dipole is the basic unit of magnetism.
Unlike electric charge, where an isolated positive or negative charge can exist, no isolated magnetic 'monopole' has ever been observed. If a bar magnet is cut into pieces, each piece becomes a smaller magnet with its own north and south pole — the poles can never be separated. This means the fundamental, indivisible unit of magne …
- CBSE 2025Set ANNUAL1 markQ.The strength of bar magnet is maximum at its center. (T/F)
›Reveal solutionSolution
This statement is False — the strength (pole strength) of a bar magnet is maximum at its two ends (poles), not at its centre.
A bar magnet's magnetism is concentrated near its two ends, called the north and south poles, where the pole strength m is maximum. At the exact centre of the magnet, the effects of the two equal and opposite poles (north and south) tend to cancel, so the net magnetic effect (and hence the 'strength' measurable there) is mini …
- CBSE 2024Set A1 markMCQQ.The value of magnetic potential at a distance r from a pole strength m is (A) (μ₀/4π)(m/r) (B) (μ₀/4π)(m/r^2) (C) (μ₀/4π)(m/r^3) (D) zero
›Reveal solutionSolution
Magnetic scalar potential of a pole falls as 1/r: V = (μ₀/4π)(m/r).
By analogy with electrostatics, a magnetic pole of strength m produces a field that falls off as 1/r2:
B=4πμ0r2m,
and a magnetic scalar potential that falls off as 1/r:
V=4πμ0rm.
…
- CBSE 2022Set I1 markMCQQ.S.I. unit of pole strength is (A) Am^-1 (B) Am^-2 (C) Am (D) Fm
›Reveal solutionSolution
SI unit of magnetic pole strength = ampere-metre (A·m).
The magnetic dipole moment of a bar magnet is m = (pole strength) × (magnetic length), i.e. M = q_m × 2l. The SI unit of magnetic moment is A·m² (same as current × area). Since the magnetic length has unit metre:
…
- CBSE 2021Set A1 markMCQQ.S.I. unit of magnetic pole strength is (A) N (B) N/A.m (C) A.m (D) A.m/N
›Reveal solutionSolution
Magnetic pole strength has SI unit ampere·metre (A·m).
Magnetic pole strength m is related to magnetic moment by M = m × 2l, where 2l is the magnetic length. Since magnetic moment has unit ampere·metre² (A·m²) and length has unit metre, pole strength has unit:
…
- CBSE 2019Set ANNUAL1 markQ.________ Law for magnetism establishes that monopoles do not exist.
›Reveal solutionSolution
Gauss's Law for magnetism states the net magnetic flux through any closed surface is always zero — the direct proof that isolated magnetic monopoles do not exist.
Gauss's law for magnetism is written as:
∮B⋅dA=0
for any closed surface. This is in contrast to Gauss's law for electric fields, ∮E⋅dA=qenc/ε0, which is non-zero whenever a net electric charge (monopole) is enclosed.
…
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