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Physics · 2024 · 3 marks
CBSE 2024 · Region 5 · Set 1 · Q24
(a) (i) State Lenz's Law. In a closed circuit, the induced current opposes the change in magnetic flux that produced it as per the law of…
(a)(i)
State Lenz's Law. In a closed circuit, the induced current opposes the change in magnetic flux that produced it as per the law of conservation of energy. Justify.
(ii)
A metal rod of length $\displaystyle 2$ m is rotated with a frequency $\displaystyle 60 \mathrm{rev} / \mathrm{s}$ about an axis passing through its centre and perpendicular to its length. A uniform magnetic field of 2T perpendicular to its plane of rotation is switched-on in the region. Calculate the e.m.f. induced between the centre and the end of the rod.
24. (b) (i) State and explain Ampere's circuital law.
(ii)
Two long straight parallel wires separated by $\displaystyle 20$ cm , carry $\displaystyle 5$ A and $\displaystyle 10$ A current respectively, in the same direction. Find the magnitude and direction of the net magnetic field at a point midway between them.
Marking-scheme solution
(i)
The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
In a closed loop, if the polarity of the induced emf were such that the induced current favours the change in magnetic flux, then the magnetic flux and consequently the current would go on increasing without any external source of energy. This violates the law of conservation of energy.
(ii)
$\displaystyle \varepsilon=\frac{1}{2} B l^{2} \omega$
$\displaystyle =\frac{1}{2} \times 2 \times(2)^{2} \times(2 \pi \times 60)$
$\displaystyle =480 \pi \mathrm{~V}$
$\displaystyle =1.51 \times 10^{3} \mathrm{~V}$
(i)
Line integral of magnetic field over a closed loop in vacuum is equal to $\displaystyle \mu_{0}$ times the total current passing through the loop.
$\displaystyle \oint \vec{B} \cdot d \vec{l}=\mu_{0} I$
The integral in this expression is over a closed loop coinciding with the boundary of the surface.
(ii)
$\displaystyle B=\frac{\mu_{0} I}{2 \pi r}$
Net magnetic field $\displaystyle B=B_{2}-B_{1}$
$\displaystyle B=\frac{\mu_{0} \times 10^{2}}{20 \pi}[10-5]$
$\displaystyle B=\frac{4 \pi \times 10^{-7} \times 10^{2} \times 5}{20 \pi}$
$\displaystyle B=10^{-5} \mathrm{~T}$
Along the direction of the magnetic field produced by the conductor carrying current $\displaystyle 10$ A.
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