2020 A/L Physics Past Paper · Question 40 (English Medium)

Period of revolution of a positive charge circling an endless charged wire

An endless thin straight wire lying on the zz-axis has a uniform linear charge density −λ-\lambda. A small positive charge +q+q of mass mm is allowed to move round the wire on a circular path of radius rr in the xyxy-plane. Which of the following gives the period of revolution of the charge?

  1. (1) 8π3r2mε0λq\sqrt{\dfrac{8\pi^3 r^2 m\varepsilon_0}{\lambda q}}
  2. (2) 4π2r3mε0λq\sqrt{\dfrac{4\pi^2 r^3 m\varepsilon_0}{\lambda q}}
  3. (3) λq8π3r2mε0\sqrt{\dfrac{\lambda q}{8\pi^3 r^2 m\varepsilon_0}}
  4. (4) λq4π2r3mε0\sqrt{\dfrac{\lambda q}{4\pi^2 r^3 m\varepsilon_0}}
  5. (5) 8r2mλε0q\sqrt{\dfrac{8 r^2 m\lambda}{\varepsilon_0 q}}
Answer and explanation

Answer: (1) 8π3r2mε0λq\sqrt{\dfrac{8\pi^3 r^2 m\varepsilon_0}{\lambda q}}

What the question asks

We must find: the period of revolution TT of +q+q (mass mm) circling the −λ-\lambda wire on a circle of radius rr — in symbols.

Relevant theory

Like charges (+, + and -, -) repel each other while unlike charges (+, -) attract each other.

Consider a cylindered Gaussean surface of which the axis is the axis of the wire and radius is r.

Since a body is moving with an acceleration only under the action of an external force, a body performing circular motion is doing so under the action of an external force acting in the direction of the centripetal acceleration. This force is thus known as "centripetal force" and is given by.

F=mv2r=mrω2F = \frac{mv^2}{r} = mr\omega^2

Period of rotation is the time taken for one complete rotation.

Required equations

E×2πrl=λlε0E \times 2\pi r l = \frac{\lambda l}{\varepsilon_0}

F=mv2rF = \frac{mv^2}{r}

T=2πrvT = \frac{2\pi r}{v}

Reasoning and calculation

Symbols: λ\lambda — the magnitude of charge per unit length of the wire (it is negative, so +q+q is attracted towards it).

rr — the distance of the charge from the wire (the radius of its path).

vv — the speed of the charge.

EE — the electric field intensity of the wire at rr, directed towards the wire.

E=λ2πε0rE = \frac{\lambda}{2\pi\varepsilon_0 r}

qE=mv2rqE = \frac{mv^2}{r}

qλ2πε0r=mv2r\frac{q\lambda}{2\pi\varepsilon_0 r} = \frac{mv^2}{r}

v2=qλ2πε0mv^2 = \frac{q\lambda}{2\pi\varepsilon_0 m}

T=2πrv=2πr2πε0mλqT = \frac{2\pi r}{v} = 2\pi r\sqrt{\frac{2\pi\varepsilon_0 m}{\lambda q}}

T=4π2r2×2πε0mλqT = \sqrt{\frac{4\pi^2 r^2 \times 2\pi\varepsilon_0 m}{\lambda q}}

T=8π3r2mε0λqT = \sqrt{\frac{8\pi^3 r^2 m\varepsilon_0}{\lambda q}}

Reading the options

This matches 8π3r2mε0λq\sqrt{\dfrac{8\pi^3 r^2 m\varepsilon_0}{\lambda q}} — option (1).

(2), (4) carry an r3r^3 from a point charge's field; (3), (4) are inverted; (5) has λ\lambda in the numerator.

Checks

sqrt(8∗pi3∗r2∗m∗e0/(lam∗q))sqrt(8 * pi^3 * r^2 * m * e0 / (lam * q))

Dimension check: the unit of r2mε0λq\dfrac{r^2 m\varepsilon_0}{\lambda q} is m2⋅kg⋅C2N−1m−2C m−1⋅C=kg mN=s2\dfrac{\text{m}^2 \cdot \text{kg} \cdot \text{C}^2\text{N}^{-1}\text{m}^{-2}}{\text{C m}^{-1} \cdot \text{C}} = \dfrac{\text{kg m}}{\text{N}} = \text{s}^2; the root is s ✓.

Boundary limit: if λ→0\lambda \to 0 there is no attraction, so T→∞T \to \infty — the charge would not circle at all ✓ (option (5) wrongly gives T→0T \to 0 here).

If rr doubles, vv stays the same, so TT doubles ✓.

Conclusion

T=8π3r2mε0λqT = \sqrt{\frac{8\pi^3 r^2 m\varepsilon_0}{\lambda q}}

E=λ2πε0rE = \dfrac{\lambda}{2\pi\varepsilon_0 r}, qE=mv2rqE = \dfrac{mv^2}{r} ⇒ T=8π3r2mε0λqT = \sqrt{\dfrac{8\pi^3 r^2 m\varepsilon_0}{\lambda q}} — the answer is (1).

Using a point charge's field kQr2\dfrac{kQ}{r^2} for the wire's field.

By Gauss's theorem the field of an endless wire is λ2πε0r\dfrac{\lambda}{2\pi\varepsilon_0 r}; it falls as 1r\dfrac{1}{r}.

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