Gravitation and nuclear fusion

Electricity and magnetism

Chemical Bonding and Planetary motion

None

The study of attraction or repulsion between two magnets is called electromagnetism

The study of electric effects due to magnetic interaction is called electromagnetism

The study of magnetic effects produced due to electric current is called electromagnetism

The study of magnetic effects produced due to electric charge is called electromagnetism

Magentism

Gravitation

Nuclear Fission

Chemical Bonding

John Ambrose Fleming

Hans Christian Oersted

Michael Faraday

André-Marie Ampère

Left hand Thumb Rule

Right Hand Thumb Rule

By rotating the conductor

Using sonometer

downward

Upward

north

south

Neither electric field nor magnetic field

Electro-static field only

Magnetic field only

Both magnetic and electro-static field

circular lines

straight lines

concentric lines

elliptical lines

Farad

Weber

${ C }^{ 2 }{ N }^{ -1 }{ m }^{ -2 }$

${ C }^{ 2 }{ N }^{ -1 }{ m }^{ -1 }$

the direction of field $\bar{B}$ will be same as the direction of velocity $\bar{v}$

the direction of field $\bar{B}$ will be opposite as the direction of velocity $\bar{v}$

the direction of field $\bar{B}$ will be perpendicular as the direction of velocity $\bar{v}$

the direction of field $\bar{B}$ does not depend upon the direction of velocity $\bar{v}$

$\bar B \perp \bar v$

$\bar B \parallel \bar v$

it obeys inverse cube law.

it is along the line joining the electron and point of observation.

$ - \frac{{2{v _0}}}{{{B _0}\alpha }}\,\;$

$\frac{{{v _0}}}{{{B _0}\alpha }}\;$

$\;\frac{{2{v _0}}}{{{B _0}\alpha }}$

$ - \frac{{{v _0}}}{{{B _0}\alpha }}$

greater than at Q

same as at Q

less than at Q

greater or less than at Q depending upon the strength of current

>X

<X

=X

all

the current passing through it.

the voltage across it

the power through it

all

>X

<X

=X

all

$\vec { dB } =\dfrac { { \mu } _{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ r } $

$\vec { dB } =\dfrac { { \mu } _{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \hat { r } }{ { r }^{ 3 } } $

$\vec { dB } =\dfrac { { \mu } _{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ { r }^{ 3 } } $

$\vec { dB } =\dfrac { { \mu } _{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ { r }^{ 4 } } $

$\dfrac{\mu _0 il}{2\pi R}$

$\dfrac{8\mu _0l}{\pi R}$

$\dfrac{\mu _0l}{4\pi R}$

$\dfrac{2\mu _0l}{\pi R}$

Electric current

magnetic effect

heating effect

all of the above

1:1

2:1

1:2

1:4

$\dfrac {\mu _0i}{4\pi} \dfrac {\vec {dl}\times \vec r}{r^3}$

$\dfrac {\mu _0i}{4\pi} \dfrac {\vec r\times \vec {dl}}{r^3}$

$-\dfrac {\mu _0i}{4\pi} \dfrac {\vec r\times \vec {dl}}{r^3}$

$-\dfrac {\mu _0i}{4\pi} \dfrac {\vec {dl}\times \vec r}{r^3}$

$ d\overrightarrow { B } =\dfrac { \mu _ o }{ 4\pi } \dfrac { di\left( \overrightarrow { l } \times \overrightarrow { r } \right) }{ r^ 2 } $

$ d\overrightarrow { B } =\dfrac { \mu _ o }{ 4\pi } \dfrac { i\left( \overrightarrow { dl } \times \overrightarrow { r } \right) }{ r^ 2 } $

$ d\overrightarrow { B } =\dfrac { \mu _ o }{ 4\pi } \dfrac { i\left( \overrightarrow { r } \times \overrightarrow { dl } \right) }{ r^ 2 } $

$ d\overrightarrow { B } =\dfrac { \mu _ o }{ 4\pi } \dfrac { i\left( \overrightarrow { dl } \times \overrightarrow { r } \right) }{ r^ 3 } $

$He^{}$

Proton

$\alpha-particle$

$Li^{++}$

iron rod

moving charge

moving magnet

stationary charge

$ \frac {i\triangle l sin \theta}{r^2} $

$ \frac {\mu _o}{4 \pi} \frac {i \triangle l sin \theta}{r} $

$ \frac {\mu _o}{4\pi} \frac {i \triangle l sin \theta}{r^2} $

$ \frac {\mu _o}{4 \pi} i \triangle l sin \theta $

$\displaystyle dB = \frac{\mu _0 i}{4 \pi} \left ( \frac{\vec{dl} \times \vec r}{r} \right )$

$\displaystyle dB = \frac{\mu _0 i}{4 \pi} i^2 \left ( \frac{\vec{dl} \times \vec r}{r^2} \right )$

$\displaystyle dB = \frac{\mu _0 i}{4 \pi} i^3 \left ( \frac{\vec{dl} \times \vec r}{2r^2} \right )$

$\displaystyle dB = \frac{\mu _0}{4 \pi} i \left ( \frac{\vec{dl} \times \vec r}{r^3} \right )$

$\displaystyle \left ( \frac{Mv^2}{R} \right ) 2 \pi R$

$zero$

$BQ2 \pi R$

$BQv2 \pi R$

current through it

distance from it

its length

nature of meterial

$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i\left ( \dfrac{\overline{dl}\times \bar{r}}{r} \right )$

$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i^{2}\left ( \dfrac{\overline{dl}\times \bar{r}}{r^{2}} \right )$

$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i^{2}\left ( \dfrac{\overline{dl}\times \bar{r}}{r} \right )$

$\overline{dB}=\dfrac{\mu _{0}}{4\pi }i\left ( \dfrac{\overline{dl}\times \bar{r}}{r^{3}} \right )$

maximum

infinity

zero

finite value

$\dfrac{3\pi}{ 4}$

$\dfrac{\pi }{4}$

$\dfrac{\pi} {2}$

$2\pi $

$8\times {10}^{-15}N$

${10}^{4}N$

$1.6\times {10}^{-19}N$

Zero

amount of current flowing through the conductor

amount of voltage supplied to the conductor

size of conductor

shape of the conductor

I

$I^2$

$I^3$

$\sqrt{I}$

True

False

Zero

Maximum

Inversely proportional to the current

None of the above

Blonde-Rey law

Biot-Savart's law

Beer-Lambert law

Ampere's law

$\dfrac{\mu _i}{3\sqrt{3}\pi a}$

$\dfrac{3\mu _i}{2\pi a}$

$\dfrac{5\sqrt{2}\mu _i}{3\pi a}$

$\dfrac{9\mu _i}{2\pi a}$