CBSE 2024 · Region 2 · Set 1 · Q25 · 2 marks
Two long, straight, parallel conductors carry steady currents in opposite directions. Explain the nature of the force of interaction between them. Obtain an expression for the magnitude of the force between the two conductors. Hence define one ampere.Obtain an expression for the torque $\displaystyle \vec{\tau}$ acting on a current carrying loop in a uniform magnetic field $\displaystyle \overrightarrow{\mathrm{B}}$. Draw the necessary diagram.
Two long, straight, parallel conductors carry steady currents in opposite directions. Explain the nature of the force of interaction between them. Obtain an expression for the magnitude of the force between the two conductors. Hence define one ampere.
Obtain an expression for the torque $\displaystyle \vec{\tau}$ acting on a current carrying loop in a uniform magnetic field $\displaystyle \overrightarrow{\mathrm{B}}$. Draw the necessary diagram.
Marking-scheme solution
Nature of force is repulsive.
Magnetic field due to current $\displaystyle I_{a}$ at all points of conductor b:
$\displaystyle B_{a b}=\frac{\mu_{0} I_{a}}{2 \pi d}$ directed downwards
Force experienced by conductor b on its segment of length l:
$\displaystyle F_{a b}=I_{b} l B_{a b}$
$\displaystyle =\frac{\mu_{0} I_{a} I_{b}}{2 \pi d} l$ directed towards left
Similarly, force experienced by conductor a on its segment of length l:
$\displaystyle F_{b a}=\frac{\mu_{0} I_{a} I_{b}}{2 \pi d} l$ directed towards right
One ampere is that steady current which, when maintained in each of two very long straight parallel conductors of negligible cross-section placed one metre apart in vacuum, produces a force of $\displaystyle 2 \times 10^{-7}$ N/m on each conductor.
Forces on arms BC and DA are equal and opposite and act along the axis of the coil. Being collinear they cancel each other.
Forces on arms AB and CD are equal and opposite but not collinear. They form a couple.
$\displaystyle F_{1}=F_{2}=I b B$
$\displaystyle \tau=F_{1} \frac{a}{2} \sin \theta+F_{2} \frac{a}{2} \sin \theta$
$\displaystyle \tau=I a b B \sin \theta$
$\displaystyle \tau=I A B \sin \theta$ (where $\displaystyle A=a b$ & $\displaystyle m=I A$)
$\displaystyle \vec{\tau}=\vec{m} \times \vec{B}$
Moving Charges and MagnetismForce between Two Parallel Currents, the AmpereApplyvery_short_answermedium
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.