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Magnetism and Matter - The Bar Magnet

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A bar magnet is a rectangular object that possesses a magnetic field and has two poles: North (NN) and South (SS).

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Magnetic field lines form continuous closed loops, emerging from the North pole and entering the South pole outside the magnet, and moving from South to North inside.

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The Magnetic Dipole Moment M⃗\vec{M} of a bar magnet is defined as the product of its pole strength mm and the magnetic length 2l2l, directed from SS to NN: M⃗=m×2l⃗\vec{M} = m \times 2\vec{l}.

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The magnetic field strength at a point on the axial line of a short bar magnet (r≫lr \gg l) is given by Baxial=μ04π2Mr3B_{axial} = \frac{\mu_0}{4\pi} \frac{2M}{r^3}.

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The magnetic field strength at a point on the equatorial line of a short bar magnet is Bequatorial=μ04πMr3B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M}{r^3}.

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Gauss's Law for Magnetism states that the net magnetic flux through any closed surface is zero: ∮B⃗⋅dS⃗=0\oint \vec{B} \cdot d\vec{S} = 0. This implies that magnetic monopoles do not exist.

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A bar magnet placed in a uniform magnetic field B⃗\vec{B} experiences a torque τ⃗=M⃗×B⃗\vec{\tau} = \vec{M} \times \vec{B}. It does not experience any net force in a uniform field.

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The potential energy UU of a magnetic dipole in a uniform magnetic field is U=−M⃗⋅B⃗=−MBcos⁡θU = -\vec{M} \cdot \vec{B} = -MB \cos \theta.

📐Formulae

M=m×2lM = m \times 2l

Baxial=μ04π2Mr(r2−l2)2≈μ04π2Mr3 (for r≫l)B_{axial} = \frac{\mu_0}{4\pi} \frac{2Mr}{(r^2 - l^2)^2} \approx \frac{\mu_0}{4\pi} \frac{2M}{r^3} \text{ (for } r \gg l)

Bequatorial=μ04πM(r2+l2)3/2≈μ04πMr3 (for r≫l)B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M}{(r^2 + l^2)^{3/2}} \approx \frac{\mu_0}{4\pi} \frac{M}{r^3} \text{ (for } r \gg l)

τ⃗=M⃗×B⃗=MBsin⁡θ\vec{\tau} = \vec{M} \times \vec{B} = MB \sin \theta

U=−M⃗⋅B⃗=−MBcos⁡θU = -\vec{M} \cdot \vec{B} = -MB \cos \theta

W=MB(cos⁡θ1−cos⁡θ2)W = MB(\cos \theta_1 - \cos \theta_2)

∮B⃗⋅dS⃗=0\oint \vec{B} \cdot d\vec{S} = 0

💡Examples

Problem 1:

A short bar magnet has a magnetic moment of 0.48 J T−10.48 \text{ J T}^{-1}. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm10 \text{ cm} from the centre of the magnet on (a) the axis, (b) the equatorial lines (perpendicular bisector) of the magnet.

Solution:

Given: M=0.48 J T−1M = 0.48 \text{ J T}^{-1}, r=10 cm=0.1 mr = 10 \text{ cm} = 0.1 \text{ m}. Using the short magnet approximations: (a) For axial point: B=μ04π2Mr3=10−7×2×0.48(0.1)3=0.96×10−4 TB = \frac{\mu_0}{4\pi} \frac{2M}{r^3} = 10^{-7} \times \frac{2 \times 0.48}{(0.1)^3} = 0.96 \times 10^{-4} \text{ T}. Direction is along the magnet's dipole moment (S to N). (b) For equatorial point: B=μ04πMr3=10−7×0.48(0.1)3=0.48×10−4 TB = \frac{\mu_0}{4\pi} \frac{M}{r^3} = 10^{-7} \times \frac{0.48}{(0.1)^3} = 0.48 \times 10^{-4} \text{ T}. Direction is opposite to the dipole moment (N to S).

Explanation:

Calculates magnetic field using the inverse cube law for dipoles at axial and equatorial positions.

Problem 2:

A bar magnet of magnetic moment 1.5 J T−11.5 \text{ J T}^{-1} lies aligned with the direction of a uniform magnetic field of 0.22 T0.22 \text{ T}. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction?

Solution:

Initial angle θ1=0∘\theta_1 = 0^\circ, Final angle θ2=90∘\theta_2 = 90^\circ. Work done W=MB(cos⁡θ1−cos⁡θ2)W = MB(\cos \theta_1 - \cos \theta_2). M=1.5M = 1.5, B=0.22B = 0.22. W=1.5×0.22×(cos⁡0∘−cos⁡90∘)W = 1.5 \times 0.22 \times (\cos 0^\circ - \cos 90^\circ) W=0.33×(1−0)=0.33 JW = 0.33 \times (1 - 0) = 0.33 \text{ J}

Explanation:

Work done is calculated using the change in potential energy of the dipole in the external magnetic field.