2 Marks Questions - Physics STD 12 Science Questions - Vidyadip

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Question 12 Marks

Figure shows a conducting square loop placed parallel to the pole faces of a ring magnet. The Pole-faces have an area of 1cm² each and the field between the poles is 0.10T. The wires making the loop are all outside the magnetic field. If the magnet is removed in 1.0s, what is the average emf induced in the loop?

Answer

$\text{B}=0.10\text{T}$
$\text{A}=1\text{cm}^2=10^{-4}\text{m}^2$
$\text{T}=1\text{s}$
$\phi=\text{B}.\text{A=}10^{-1}\times10^{-4}=10^{-5}$
$\text{e}=\frac{\text{d}\phi}{\text{dt}}=\frac{10^{-5}}{1}=10^{-5}=10\mu\text{V}$

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Question 22 Marks

Figure shows a circular wheel of radius 10.0cm whose upper half, shown dark in the figure, is made of iron and the lower half of wood. The two junctions are joined by an iron rod. A uniform magnetic field B of magnitude 2.00 × 10^-4 T exists in the space above the central line as suggested by the figure. The wheel is set into pure rolling on the horizontal surface. If it takes 2.00 seconds for the iron part to come down and the wooden part to go up, find the average emf induced during this period.

Answer

$\phi_{1}=\text{BA},\phi_{2}=0$
$=\frac{2\times10^{-4}\times\pi(0.1)^2}{2}=\pi\times10^{-5}$
$\text{E}=\frac{\text{d}\phi}{\text{dt}}=\frac{\pi\times10^{-6}}{2}=1.57\times10^{-6}\text{V}$

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Question 32 Marks

Find the mutual inductance between the circular coil and the loop shown in figure.

Answer

emf induced $=\frac{\pi\mu_0\text{Na}^2\text{a}'^2\text{ERV}}{2\text{L}(\text{a}^2+\text{x}^2)^{\frac{3}{2}}\Big(\frac{\text{R}}{\text{L}\text{x}}+\text{r}\Big)^2}$
$\frac{\text{dl}}{\text{dt}}=\frac{\text{ERV}}{\text{L}\Big(\frac{\text{Rx}}{\text{L}}+\text{r}\Big)^2}$
$\mu=\frac{\text{E}}{\frac{\text{di}}{\text{dt}}}=\frac{\text{N}\mu_0\pi\text{a}^2\text{a}'^2}{2(\text{a}^2+\text{x}^2)^{\frac{3}{2}}}$

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Question 42 Marks

An inductor is connected to a battery through a switch. Explain why the emf induced in the inductor is much larger when the switch is opened as compared to the emf induced when the switch is closed.

Answer

Due to self inductance the emf produced when circuit is open the growth of current in inductor is small but when switch is closed the decay in current is faster thus emf induced is more.

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Question 52 Marks

The two rails of a railway track, insulated from each other and from the ground, are connected to a millivoltmeter. What will be the reading of the millivoltmeter when a train travels on the track at a speed of 180km/h^-1? The vertical component of earth's magnetic field is 0.2 × 10^-4 T and the rails are separated by 1m.

Answer

v = 180km/h = 50m/s
B = 0.2 × 10^-4T, L = 1m
E = Bvl = 0.2I 10^-4 × 50 = 10^-3V
$\therefore$ The voltmeter will record 1mv.

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Question 62 Marks

A wire-loop confined in a plane is rotated in its own plane with some angular velocity. A uniform magnetic field exists in the region. Find the emf induced in the loop.

Answer

$\text{i}=\text{v}(\text{B}\times\text{l})$
$=\text{vBl}\cos\theta$
$\theta$ is angle between normal to plane and $\vec{\text{B}}=90^{\circ}$
$=\text{vBl}\cos90^{\circ}=0.$

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Question 72 Marks

The north pole of a magnet is brought down along the axis of a horizontal circular coil. As a result, the flux through the coil changes from 0.35 weber to 0.86 weber in an interval of half a second. Find the average emf induced during this period. Is the induced current clockwise or anticlockwise as you look into the coil from the side of the magnet?

Answer

$\phi_1=0.35 \ \text{weber}, \ \phi_2=0.85 \ \text{weber}$
$\text{D}\phi=\phi_2=(0.85-0.35) \ \text{weber}=0.5 \ \text{weber}$
$\text{dt}=0.5\text{sec}$
$\text{E}=\frac{\text{d}\phi}{\text{dt}'}=\frac{0.5}{0.5}=1\text{V}$
The induced current is anticlockwise as seen from above.

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Question 82 Marks

Figure shows a long U-shaped wire of width l placed in a perpendicular magnetic field B. A wire of length l is slid on the U-shaped wire with a constant velocity v towards right. The resistance of all the wires is r per unit length. At t = 0, the sliding wire is close to the left edge of the U-shaped wire. Draw an equivalent circuit diagram, showing the induced emf as a battery. Calculate the current in the circuit.

Answer

$\text{E}=\text{Bvl}$
Resistance = r × total length
$=\text{r}\times2(\text{l}+\text{vt})=24(\text{l}+\text{vt})$
$\text{i}=\frac{\text{Bvl}}{2\text{r}(\text{l}+\text{vt})}$

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Question 92 Marks

Calculate the dimensions of:

$\int\vec{\text{E}}.\vec{\text{dl}}.$
$\text{vBl}$
$\frac{\text{d}\psi_{\text{B}}}{\text{dt}}$

The symbols have their usual meanings.

Answer

$\int\vec{\text{E}}.\vec{\text{dl}}=\text{ML}\text{T}^{-3}\text{l}^{-1}=\text{ML}^2\text{l}^{-1}\text{T}^{-3}$
$\text{vBl}=\text{LT}^{-1}\times\text{Ml}^{-1}\text{T}^{-2}\times\text{L}=\text{ML}^2\text{l}^{-1}\text{T}^{-3}$
$\frac{\text{d}\phi_{\text{B}}}{\text{dt}}=\text{Ml}^{-1}\text{T}^{-2}\times\text{L}^2=\text{ML}^2\text{l}^{-1}\text{T}^{-2}$

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Question 102 Marks

Figure shows a conducting circular loop of radius a placed in a uniform, perpendicular magnetic field B. A thick metal rod OA is pivoted at the centre O. The other end of the rod touches the loop at A. The centre O and a fixed point C on the loop are connected by a wire OC of resistance R. A force is applied at the middle point of the rod OA perpendicularly, so that the rod rotates clockwise at a uniform angular velocity $\omega.$ Find the force.

Answer

$\text{e}=\text{Bvl}=\frac{\text{B}\times\text{a}\times\omega\times\text{a}}{2}$

$\text{i}=\frac{\text{Ba}^2\omega}{2\text{R}}$

$\text{F}=\text{ilB}=\frac{\text{B}\text{a}^2\omega}{2\text{R}}\times\text{a}\times\text{B}=\frac{\text{B}^2\text{a}^2\omega}{2\text{R}}$ towards right of OA.

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Question 112 Marks

A conducting disc of radius r rotates with a small but constant angular velocity $\omega$ about its axis. A uniform magnetic field B exists parallel to the axis of rotation. Find the motional emf between the centre and the periphery of the disc.

Answer

V at a distance $\frac{\text{r}}{2}$

From the centre $=\frac{\text{r}\omega}{2}$

$\text{E}=\text{BlV}$

$\Rightarrow\text{E}=\text{B}\times\text{r}\times\frac{\text{r}\omega}{2}=\frac{1}{2}\text{B}\text{r}^2\omega$

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Question 122 Marks

Consider a small cube of volume 1mm³ at the centre of a circular loop of radius 10cm carrying a current of 4A. Find the magnetic energy stored inside the cube.

Answer

Energy density $=\frac{\text{B}^2}{2\mu_0}$

Total energy stored $=\frac{\text{B}^2\text{V}}{2\mu_0}=\frac{\big(\frac{\mu_0\text{i}}{2\text{r}}\big)^2}{2\mu_0}\text{V}=\frac{\mu_0\text{i}^2}{4\text{r}^2\times2}\text{V}$

$=\frac{4\pi\times10^{-7}\times4^2\times1\times10^{-9}}{4\times(10^{-1})^2\times2}=8\pi\times10^{-14} \text{J}.$

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Question 132 Marks

The current generator I_g shown in figure, sends a constant current i through the circuit. The wire ab has a length l and mass m and can slide on the smooth, horizontal rails connected to I_g. The entire system lies in a vertical magnetic field B. Find the velocity of the wire as a function of time.

Answer

Force on the wire $=\text{ilB}$
Acceleration $=\frac{\text{ilB}}{\text{m}}$
Velocity $=\frac{\text{ilBt}}{\text{m}}$

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Question 142 Marks

A conducting circular loop having a radius of 5.0cm, is placed perpendicular to a magnetic field of 0.50T. It is removed from the field in 0.50s. Find the average emf produced in the loop during this time.

Answer

$\phi_1=\text{BA}=0.5\times\pi(5\times10^{-2})^2=5\pi25\times10^{-5}=125\times10^{-5}$
$\phi_2=0$
$\text{E}=\frac{\phi_1-\phi_2}{\text{t}}=\frac{125\pi\times10^{-5}}{5\times10^{-1}}=25\pi\times10^{-4}=7.8\times10^{-3}$

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Question 152 Marks

An inductor-coil of inductance 17mH is constructed from r wire of length 100m and cross-sectional area a copper wire of length 100m and cross-sectional area 1mm². Calculate the time constant of the circuit if this inductor is joined across an ideal battery. The resistivity of copper $=1.7\times10^{-8}\Omega-\text{m}.$

Answer

$\text{L}=17\text{mH}, \ \text{l}=100\text{m}, \ \text{A}=1\text{mm}^2$
$=1\times10^{-6}\text{m}^2, \ \text{f}_{\text{cu}}=1.7\times10^{-8}\Omega-\text{m}$$$
$\text{R}=\frac{\text{f}_\text{cu}\text{l}}{\text{A}}=\frac{1.7\times10^{-8}\times100}{1\times10^{-6}}=1.7 \Omega$
$\text{i}=\frac{\text{L}}{\text{R}}=\frac{0.17\times10^{-8}}{1.7}=10^{-2} \text{ sec}=10 \ \text{m} \ \text{sec}.$

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Question 162 Marks

A pivoted aluminium bar falls much more slowly through a small region containing a magnetic field than a similar bar of an insulating material. Explain.

Answer

Aluminum rod while falling will experience magnetic field and thus it will induce some eddy currents in it and thus it will feel some attraction towards magnetic field and thus it will fall slow as compared to insulating material.

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Question 172 Marks

A conducting circular loop of area 1mm² is placed coplanarly with a long, straight wire at a distance of 20cm from it. The straight wire carries an electric current which changes from 10A to zero in 0.1s. Find the average emf induced in the loop in 0.1s.

Answer

$\text{A}=1\text{mm}^2; \ \text{i}=10\text{A}, \ \text{d}=20\text{cm}; \ \text{dt}=0.1\text{s}$
$\text{e}=\frac{\text{d}\phi}{\text{dt}}=\frac{\text{BA}}{\text{dt}}=\frac{\mu_0\text{i}}{2\pi\text{d}}\times\frac{\text{A}}{\text{dt}}$
$=\frac{4\pi\times10^{-7}\times10}{2\pi\times2\times10^{-1}}\times\frac{10^{-6}}{1\times10^{-1}}=1\times10^{-10}\text{V}$

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Question 182 Marks

A copper wire bent in the shape of a semicircle of radius r translates in its plane with a constant velocity v. A uniform magnetic field B exists in the direction perpendicular to the plane of the wire. Find the emf induced between the ends of the wire if:

The velocity is perpendicular to the diameter joining free ends.
The velocity is parallel to this diameter.

Answer

Component of length moving perpendicular to V is 2R

$\therefore$ E = Bv2R

Component of length perpendicular to velocity = 0

$\therefore$ E = 0

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Question 192 Marks

The coil of a moving-coil galvanometer keeps on oscillating for a long time if it is deflected and released. If the ends of the coil are connected together, the oscillation stops at once. Explain

Answer

When the ends are not connected the coil acts as inductor thus it undergoes inductance and keeps varying the current and keeps oscillating. But when ends are connected the coil is close loop thus there is no inductance.

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Question 202 Marks

Figure shows a wire sliding on two parallel, conducting rails placed at a separation l. A magnetic field B exists in a direction perpendicular to the plane of the rails. What force is necessary to keep the wire moving at a constant velocity v?

Answer

F_magnetic = ilB
This force produces an acceleration of the wire.
But since the velocity is given to be constant.
Hence net force acting on the wire must be zero.

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Question 212 Marks

The mutual inductance between two coils is 2.5H. If the current in one coil is changed at the rate of 1A/s, what will be the emf induced in the other coil?

Answer

$\text{M}=2.5\text{H}$
$\frac{\text{dl}}{\text{dt}}=\frac{\text{lA}}{\text{s}}$
$\text{E}=-\mu\frac{\text{dl}}{\text{dt}}$
$\Rightarrow\text{E}=2.5\times1=2.5\text{V}$

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Question 222 Marks

A conducting loop of face-area A and resistance R is placed perpendicular to a magnetic field B. The loop is withdrawn completely from the field. Find the charge which flows through any cross-section of the wire in the process. Note that it is independent of the shape of the loop as well as the way it is withdrawn.

Answer

Area = A, Resistance = R, B = Magnetic field
$\phi=\text{BA}=\text{Ba}\cos0^{\circ}=\text{BA}$
$\text{e}=\frac{\text{d}\phi}{\text{dt}}=\frac{\text{BA}}{1};\text{i}=\frac {\text{e}}{\text{R}}=\frac{\text{BA}}{\text{R}}$
$\phi=\text{iT}=\frac{\text{BA}}{\text{R}}$

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Question 232 Marks

Metallic (nonferromagnetic) and nonmetallic particles in a solid waste may be separated as follows. The waste is allowed to slide down an incline over permanent magnets. The metallic particles slow down as compared to the norunetallic ones and hence are separated. Discuss the role of eddy currents in the process.

Answer

As the metallic particle slide over permanent magnet they are attracted towards magnet due to formation of eddy current on the metallic body and thus they slow down and are separated from rest of material.

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Question 242 Marks

The battery discussed in the previous question is suddenly disconnected. Is a current induced in the other loop? If yes, when does it start and when does it end? Do the loops attract each other or repel?

Answer

As the battery is disconnected the current starts decreasing in lop 1 and thus now direction of current is in opposite direction as it was earlier and it ends when both loops are in equilibrium.

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Question 252 Marks

An inductor-coil carries a steady-state current of 2.0A when connected across an ideal battery of emf 4.0V. If its inductance is 1.0H, find the time constant of the circuit.

Answer

$\text{i}=2\text{A},\text{E}=4\text{V},\text{L}=1\text{H}$
$\text{R}=\frac{\text{E}}{\text{i}}=\frac{4}{2}=2$
$\text{i}=\frac{\text{L}}{\text{R}}=\frac{1}{2}=0.5$

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Question 262 Marks

A 10m wide spacecraft moves through the interstellar space at a speed 3 × 10⁷m/s^-1. A magnetic field B = 3 × 10^-10T exists in the space in a direction perpendicular to the plane of motion. Treating the spacecraft as a conductor, calculate the emf induced across its width.

Answer

l = 10m, v = 3 × 10⁷m/s, B = 3 × 10^-10T
Motional emf = Bvl
= 3 × 10^-10 × 3 × 10⁷ × 10
= 9 × 10^-3= 0.09V

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2 Marks Questions - Physics STD 12 Science Questions - Vidyadip