1.
A coil of inductance $$300 mH$$ and resistance $$2\,\Omega $$ is connected to a source of voltage $$2 V.$$ The current reaches half of its steady state value in
KEY CONCEPT : The charging of inductance given by, $$i = {i_0}\left( {1 - {e^{ - \frac{{Rt}}{L}}}} \right)$$
$$\frac{{{i_0}}}{2} = {i_0}\left( {1 - {e^{ - \frac{{Rt}}{L}}}} \right) \Rightarrow {e^{\frac{{Rt}}{L}}} = \frac{1}{2}$$
Taking log on both the sides,
$$\eqalign{
& - \frac{{Rt}}{L} = \log 1 - \log 2 \cr
& \Rightarrow t = \frac{L}{R}\log 2 = \frac{{300 \times {{10}^{ - 3}}}}{2} \times 0.69 \cr
& \Rightarrow t = 0.1\,{\text{sec}} \cr} $$
2.
A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :
A.
development of air current when the plate is placed
B.
induction of electrical charge on the plate
C.
shielding of magnetic lines of force as aluminium is a paramagnetic material.
D.
electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.
Answer :
electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.
Energy stored in a current carrying coil is in the form of magnetic field. Total work done by the external source in building up current from zero to $${i_0}$$ is $$W = \frac{1}{2}Li_0^2$$
where, $$L$$ is self-inductance of coil.
4.
A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :
A.
developement of air current when the plate is placed
B.
induction of electrical charge on the plate
C.
shielding of magnetic lines of force as aluminium is a paramagnetic material.
D.
electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.
Answer :
electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.
Because of the Lenz's law of conservation of energy.
5.
A uniform magnetic field is restricted within a region of radius $$r.$$ The magnetic field changes with time at a rate $$\frac{{dB}}{{dt}}.$$ Loop 1 of radius $$R > r$$ encloses the region $$r$$ and loop 2 of radius $$R$$ is outside the region of magnetic field as shown in the figure. Then, the emf generated is
A.
zero in loop 1 and zero in loop 2
B.
$$ - \frac{{dB}}{{dt}}\pi {r^2}$$ in loop 1 and $$ - \frac{{dB}}{{dt}}\pi {r^2}$$ in loop 2
C.
$$ - \frac{{dB}}{{dt}}\pi {R^2}$$ in loop 1 and zero in loop 2
D.
$$ - \frac{{dB}}{{dt}}\pi {r^2}$$ in loop 1 and zero in loop 2
Answer :
$$ - \frac{{dB}}{{dt}}\pi {R^2}$$ in loop 1 and zero in loop 2
Induced emf in the region is given by
$$\left| e \right| = \frac{{d\phi }}{{dt}}$$
where, $$\phi = BA = \pi {r^2}B$$
$$ \Rightarrow \frac{{d\phi }}{{dt}} = - \pi {r^2}\frac{{dB}}{{dt}}$$
Rate of change of magnetic flux associated with loop 1
$${e_1} = - \frac{{d{\phi _1}}}{{dt}} = - \pi {r^2}\frac{{dB}}{{dt}}$$
Similarly $${e_2} = $$ emf associated with loop 2
$$ = - \frac{{d{\phi _2}}}{{dt}} = 0\,\,\left[ {\because {\phi _2} = 0} \right]$$
6.
A straight conducting metal wire is bent in the given shape and the loop is closed. Dimensions are as shown in the figure. Now the assembly is heated at a constant rate $$\frac{{dT}}{{dt}} = {1^ \circ }C/s.$$ The assembly is kept in a uniform magnetic field $$B = 1\,T,$$ perpendicular into the paper. Find the current in the loop at the moment, when the heating starts. Resistance of the loop is $$10\,\Omega $$ at any temperature. Coefficient of linear expansion $$\alpha = \frac{{{{10}^{ - 6}}}}{{^ \circ C}}.$$
A.
$$1.5 \times {10^{ - 6}}A\,{\text{anticlockwise}}$$
B.
$$1.5 \times {10^{ - 6}}A\,{\text{clockwise}}$$
C.
$$0.75 \times {10^{ - 6}}A\,{\text{anticlockwise}}$$
D.
$$0.75 \times {10^{ - 6}}A\,{\text{clockwise}}$$
Rate of change of area of the loop
$$\eqalign{
& \frac{{dA}}{{dt}} = A,\beta \frac{{dT}}{{dt}} = A.\left( {2\alpha } \right)\frac{{dT}}{{dt}} = \frac{3}{4} \times 2 \times {10^{ - 6}} \times 1 \cr
& = 11.5 \times {10^{ - 6}}{m^2}/s \cr
& {\text{emf}} = - \frac{{d\phi }}{{dt}} = - \frac{{\beta .dA}}{{dt}} = - 1.5 \times {10^{ - 6}}V \cr} $$
current in the loop $$ = 1.5 \times {10^{ - 6}}A$$
The direction will be anticlockwise as the induced current will try to negate the increase in fluix due to increase in area.
7.
If $$N$$ is the number of turns in a coil, the value of self-inductance varies as
Magnetic flux, $$\phi = BA$$
and magnetic field due to circular coil is $$B = \frac{{{\mu _0}Ni}}{{2R}}$$
As self-inductance, $$L = \frac{{N\phi }}{i}$$
$$\eqalign{
& \therefore L = \frac{N}{i}\left( {BA} \right) = \frac{N}{i}\left( {\frac{{{\mu _0}Ni}}{{2R}}} \right)A \cr
& = \frac{{{\mu _0}{N^2}A}}{{2R}} \cr
& \therefore L \propto {N^2} \cr} $$
8.
A resistor $$'R'$$ and $$2\mu F$$ capacitor in series is connected through a switch to $$200\,V$$ direct supply. Across the capacitor is a neon bulb that lights up at $$120\,V.$$ Calculate the value of $$R$$ to make the bulb light up $$5\,s$$ after the switch has been closed.$$\left( {{{\log }_{10}}2.5 = 0.4} \right)$$
9.
An inductor of inductance $$L = 400 mH$$ and resistors of resistance $${R_1} = 2\Omega $$ and $${R_2} = 2\Omega $$ are connected to a battery of emf $$12\,V$$ as shown in the figure. The internal resistance of the battery is negligible. The switch $$S$$ is closed at $$t = 0.$$ The potential drop across $$L$$ as a function of time is:
A.
$$\frac{{12}}{t}{e^{ - 3t}}V$$
B.
$$6\left( {1 - {e^{ - \frac{t}{{0.2}}}}} \right)V$$
Let current $${i_1}$$ in straight wire be upward. Then the magnetic field due to the straight wire has magnitude $${B_1} = \frac{{{\mu _0}{i_1}}}{{2\pi r}}$$ at distance $$r.$$ In accrodance with right hand rule, $${B_1}$$ points inward to the plane of page. We consider a differential strip of thickness $$dr,$$ area $$d{A_2} = a\,dr.$$ Magnetic flux through area $$dA,d{\phi _B} = {B_1}\left( {a\,dr} \right).$$
Total flux through the loop,
$$\eqalign{
& {\phi _B} = \int {{B_1}adr = } \int_c^{c + b} {\frac{{{\mu _0}{i_1}}}{{2\pi r}}} a\,dr \cr
& = \frac{{{\mu _0}{i_1}a}}{{2\pi }}\int_c^{c + b} {\frac{{dr}}{r}} = \frac{{{\mu _0}{i_1}a}}{{2\pi }}\ln \left( {\frac{{c + b}}{c}} \right) \cr} $$
Therefore mutual inductance,
$$M = \frac{\phi }{{{i_1}}} = \frac{{{\mu _0}a}}{{2\pi }}\ln \left( {1 + \frac{b}{c}} \right)$$