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Physics · 2026 · 5 marks
CBSE 2026 · Region 1 · Set 1 · Q33
(a) State Faraday's law of electromagnetic induction. (b) Derive an expression for the self-inductance of an air-filled long solenoid of length l and…
(a)
State Faraday's law of electromagnetic induction.
(b)
Derive an expression for the self-inductance of an air-filled long solenoid of length $\displaystyle l$ and cross-sectional area A having N turns.
(c)
A conducting rod of length $\displaystyle 50$ cm, with one end pivoted, is rotated with angular speed of $\displaystyle 60$ rpm in a uniform magnetic field of $\displaystyle 4.0$ mT directed perpendicular to the plane of rotation of rod. Find the emf induced in the rod.
(a)
Draw a labelled diagram of a step-up transformer. State the principle on which it works and obtain the ratio of secondary voltage to primary voltage in terms of number of turns and currents in the two coils.
(b)
The ratio of the number of turns in the primary to the secondary of an ideal transformer is $\displaystyle 1$ : 5. If $\displaystyle 5$ kW power at $\displaystyle 200$ V is supplied to the primary, find
(i)
current in the primary, and
(ii)
output voltage.
Marking-scheme solution
(a)
The magnitude of the induced e.m.f in a circuit is equal to the time rate of change of magnetic flux through the circuit.
Alternatively: $\displaystyle \varepsilon=-\dfrac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}$
(b)
Magnetic field due to current carrying long solenoid of length $\displaystyle l$ area of cross section A having n turns per unit length is $\displaystyle \mathrm{B}=\mu_{0} \mathrm{nI}$
Total magnetic flux linked with the solenoid is
$\displaystyle \mathrm{N} \phi_{\mathrm{B}}=(n l)\left(\mu_{0} n \mathrm{I}\right)(A)$
$\displaystyle =\mu_{0} n^{2} A l \mathrm{I}$
Where $\displaystyle n l$ is total number of turns.
$\displaystyle \therefore$ Self inductance of the solenoid
$\displaystyle \mathrm{L}=\dfrac{\mathrm{N} \phi_{\mathrm{B}}}{\mathrm{I}}$
$\displaystyle \mathrm{L}=\mu_{0} n^{2} A l$
(c)
Induced e.m.f
$\displaystyle \varepsilon=\dfrac{1}{2} \mathrm{B} l^{2} \omega$
$\displaystyle =\dfrac{1}{2} \times 4 \times 10^{-3} \times\left(50 \times 10^{-2}\right)^{2} \times(2 \pi \times 1)$
$\displaystyle =3.14 \mathrm{mV}$
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