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Physics · 2026 · 3 marks
CBSE 2026 · Region 2 · Set 1 · Q25
(i) Define mutual inductance of a pair of coils. Write its SI unit. (ii) A long solenoid of radius R and length L has n turns per unit length. A…
(i)
Define mutual inductance of a pair of coils. Write its SI unit.
(ii)
A long solenoid of radius R and length L has n turns per unit length. A circular loop of radius $\displaystyle \mathrm{r}(<\mathrm{R})$ is placed inside at the centre of the solenoid such that its axis coincides with the axis of the solenoid. Obtain the mutual inductance of the solenoid and the loop.
OR Two long straight parallel conductors A and B carrying steady currents $\displaystyle \mathrm{I}_{\mathrm{a}}$ and $\displaystyle \mathrm{I}_{\mathrm{b}}$ in the same direction are separated by a distance d. Deduce the expressions for the force acting on length L of conductor B due to conductor A and show it in figure. Write the expression for the force acting on length L of conductor A due to conductor B and show that it follows Newton's third law.
Marking-scheme solution
(i)
It is the ratio of the magnetic flux linked with a coil to the current flowing in the neighbouring coil.
Alternatively: It is the ratio of magnitude of induced emf in a coil to the rate of change of current in the neighbouring coil.
$\displaystyle M=\dfrac{\phi}{I}$ or $\displaystyle M=\dfrac{|\varepsilon|}{d i / d t}$
SI unit: henry (H) or $\displaystyle \mathrm{Tm}^{2} \mathrm{~A}^{-1}$ or $\displaystyle \mathrm{WbA}^{-1}$ or $\displaystyle \mathrm{VsA}^{-1}$
(ii)
If current I is set up through the solenoid, the magnetic field inside at its centre $\displaystyle B=\mu_{0} n I$
Flux linked with the circular loop of radius $\displaystyle r(<R)$: $\displaystyle \phi_{B}=B A=\left(\mu_{0} n I\right)\left(\pi r^{2}\right)$
Mutual inductance $\displaystyle M=\dfrac{\phi_{B}}{I}=\mu_{0} n \pi r^{2}$
Magnetic field due to current carrying conductor A at all points along conductor B: $\displaystyle \overrightarrow{\mathrm{B}}_{a}=\dfrac{\mu_{0} I_{a}}{2 \pi d}(-\hat{k})$
Force experienced by a segment of length L of conductor B: $\displaystyle \overrightarrow{\mathrm{F}}_{\mathrm{BA}}=I_{b} \vec{L} \times \overrightarrow{\mathrm{B}}_{a}=I_{b} \mathrm{~L}\left(\dfrac{\mu_{0} I_{a}}{2 \pi d}\right)[\hat{i} \times(-\hat{k})]$
$\displaystyle \overrightarrow{\mathrm{F}}_{\mathrm{BA}}=\dfrac{\mu_{0} I_{a} I_{b} \mathrm{~L}}{2 \pi \mathrm{d}} \hat{j}$
Similarly $\displaystyle \overrightarrow{\mathrm{F}}_{\mathrm{AB}}=\dfrac{\mu_{0} I_{a} I_{b} \mathrm{~L}}{2 \pi \mathrm{d}}(-\hat{j})$
$\displaystyle \overrightarrow{\mathrm{F}}_{\mathrm{BA}}=-\overrightarrow{\mathrm{F}}_{\mathrm{AB}}$, so it follows Newton's third law.
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