CBSE 2024 · Region 4 · Set 1 · Q26 · 3 marks
A circular loop of radius $\displaystyle 10$ cm carrying current of $\displaystyle 1.0$ A lies in $\displaystyle x-\mathrm{y}$ plane. A long straight wire lies in the same plane parallel to $\displaystyle x$-axis at a distance of $\displaystyle 20$ cm as shown in figure.
Find the direction and value of current that has to be maintained in the wire so that the net magnetic field at O is zero.
Marking-scheme solution
Magnetic field due to circular current loop at its centre O:
$\displaystyle B_{1}=\frac{\mu_{0} I_{1}}{2 r}$
$\displaystyle =\frac{\mu_{0} \times 1}{2 \times 0.1}$
$\displaystyle =5 \mu_{0} \mathrm{~T}$
The magnetic field $\displaystyle B_{1}$ is perpendicular to the plane of the loop and directed inwards.
Magnetic field due to long current carrying straight wire at O:
$\displaystyle B_{2}=\frac{\mu_{0} I_{2}}{2 \pi r}$
$\displaystyle B_{2}=\frac{\mu_{0} I_{2}}{2 \pi \times 0.2} \mathrm{~T}$
For the net magnetic field at O to be zero, $\displaystyle B_{1}$ should be equal and opposite to $\displaystyle B_{2}$:
$\displaystyle 5 \mu_{0}=\frac{\mu_{0} I_{2}}{0.4 \pi}$
$\displaystyle I_{2}=2 \pi \mathrm{~A}$
$\displaystyle =6.28 \mathrm{~A}$
Direction of current in the straight wire is along the +ve x axis.
Moving Charges and MagnetismForce between Two Parallel Currents, the AmpereApplyshort_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.