Tag: magnetic field lines due to current

Questions Related to magnetic field lines due to current

The Magnetic effect of current was discovered by:

  1. John Ambrose Fleming

  2. Hans Christian Oersted

  3. Michael Faraday

  4. André-Marie Ampère


Correct Option: B
Explanation:
The magnetic effect of an electric current was discovered by Hans Christian Oersted in 1820. He demonstrated the effect by plotting compass close to a long straight wire which is carrying current and noting it's deflection. And the by reversing the direction of the current, the compass needle was deflected in the opposite sense.

How can you determine direction of magnetic field lines around a current carrying conductor

  1. Left hand Thumb Rule

  2. Right Hand Thumb Rule

  3. By rotating the conductor

  4. Using sonometer


Correct Option: B
Explanation:

Direction of magnetic field lines around a current carrying conductor can be found by RIght Hand Thumb Rule.

One metal wire is kept in east-west direction.  $I$  is the current flow due west. Then, due to magnetic field  $\vec { { B } }$  of the earth on the wire is in the........ Direction.

  1. downward

  2. Upward

  3. north

  4. south


Correct Option: C
Explanation:
according to left hand rule. if current is in west direction then magnetic field will be north direction, if force is in north west direction

A loosely wound helix made of stiff wire is mounted vertically with the lower end just touching a dish of mercury when a current from the battery is started in the coil through the mercury

  1. the wire oscillates

  2. the wire continues making contact

  3. the wire breaks contact just when the current is passed

  4. the mercury will expand by heating due to passage of current


Correct Option: A
Explanation:

According to the Biot Savart's law, a current carrying conductor produces a magnetic field in its surroundings. This will cause the helix to oscillate across a mean position.

A moving charge produces

  1. Neither electric field nor magnetic field

  2. Electro-static field only

  3. Magnetic field only

  4. Both magnetic and electro-static field


Correct Option: D
Explanation:

A charged particle produces an electric field around it independent of its motion. But when it is at rest, it doesn't produce a magnetic field. 

When it starts moving, it additionally produces a magnetic field too.

SI unit of permittivity of free space is:

  1. Farad

  2. Weber

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

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


Correct Option: C
Explanation:

Coulombic force between two charges is given by

F=1/4π€ (q'.q″/r²)

From here, the SI UNIT of permittivity comes out is

= coulomb²/Newton-meter²

C²/N-m²

If an electron is moving with velocity $\bar{v}$ produces a magnetic field $\bar{B}$, then

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

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

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

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


Correct Option: C
Explanation:

According to Biot-Savart's law, the magnetic field
$\displaystyle \overset{\rightarrow}{B} = \frac{\mu _o}{4 \pi} . \frac{q (\overset{\rightarrow}{v} \times \overset{\rightarrow}{r} ) }{r^3}$
The direction of $\overset{\rightarrow}{B}$ will be along $\overset{\rightarrow}{v} \times \overset{\rightarrow}{r}$ i.e. perpendicular to the plane containing $\overset{\rightarrow}{v}$ and $\overset{\rightarrow}{r}$.

Biot-Savart law indicates that the moving electrons (velocity $\bar v$ ) produce a magnetic field $\bar B$ such that:

  1. $\bar B \perp \bar v$

  2. $\bar B \parallel \bar v$

  3. it obeys inverse cube law.

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


Correct Option: A
Explanation:

Magnetic field produced by charges moving with velocity $\bar v$, at a distance r is $ \bar B$ = $\left ( \dfrac{\mu _0}{4\pi } \right )$.q$\dfrac{\bar v \times \hat r}{r^2}$
Therefore $\bar B \perp \bar v$

A particle of charge per unit mass $\alpha$  is released from origin with a velocity $\mathop v\limits^ \to   = {v _0}\mathop i\limits^ \wedge  $ in a uniform magnetic field $\mathop B\limits^ \to   =  - {B _0}\mathop k\limits^ \wedge  $ . If the particle passes through (0,y,0) then y is
equal to

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

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

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

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


Correct Option: C