In the post dated 2^{nd} July 2007, two questions on magnetism were given without solution. These questions with solution are given below as promised in the post:

**(1) A thin, non-conducting rod of length ‘L’ is uniformly charged to have a linear charge density ‘λ’ from its mid point to one end (The other half of the rod is uncharged). The rod is rotated about an axis perpendicular to its length and passing through the other end with angular velocity ‘ω’. The magnetic dipole moment of the rotating rod is**

**(a) zero (b) λωL ^{3}/3 (c) λωL^{3}/6 **

The magnetic dipole moment of a current loop is iA where ‘i’ is the current and ‘A’ is the area of the loop. The rotation of the rod OP with linear charge density λ from its middle to the end P is equivalent to many current loops of varying area. Considering the current loop made by the charge λdr on a small length dr of the rod at distance ‘r’ from the end O, the magnetic moment contributed by it is (λdr/T)×πr^{2} where T is the period of rotation of the rod.

Since T = 2π/ω, the above magnetic moment is ½ (λωr^{2}dr).

The magnetic dipole moment (m) due to the rotation of the entire rod is obtained by integrating the above between the ** limits L/2 and L**.

Therefore, m = ∫[½(λωr^{2}dr) = ½(λω)×[L^{3}/3 – (L/2)^{3}/3] = **7λωL ^{3}/48.**

**(2) A magnetic needle oscillates in a horizontal plane with a period ‘T’ at a place where the angle of dip is 30º. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be**

**(a) T (b) T/√3 (c) T****×****√3 (d) T****×****3 ^{1/4} (e) T**

**×**

**3**

^{–}

^{1/4}

When the magnetic needle is made to oscillate in the horizontal plane, the restoring force required for the motion is provided by the horizontal component (B_{h}) of the earth’s magnetic field B so that we have

T = 2π√(I/mB_{h}) where ‘I’ is the moment of inertia of the needle (about the axis of angular oscillation, which is through its centre of mass and is perpendicular to its length.) and ‘m’ is its magnetic dipole moment.

When the needle is made to oscillate in the vertical plane, the restoring force required for the motion is prvided by the vertical component (B_{v}) of the earth’s magnetic field B so that we have the new period

T’ = 2π√(I/mB_{v}).

Dividing the first equation by the second, T/T’ = √(B_{v}/B_{h}) from which

T’ = T/√(B_{v}/B_{h}).

But, B_{v}/B_{h} = tanθ where ‘θ’ is the angle of dip (which is 30º here) so that

T’ = T/√(tan30º) = T/√(1/√3) = **T****×3 ^{1/4}**.

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