Tag: magnetic flux density

Questions Related to magnetic flux density

Magnetic field intensity at the centre of coil of 50 turns, radius 0.5 m and carrying a current of 2 A is 

physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

  1. $0.5 \times 10^{-5}T$

  2. $3 \times 10^{-5}T$

  3. $1.25 \times 10^{-4}T$

  4. $4 \times 10^{-5}T$

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Correct Option: C

Select the incorrect option

physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

  1. Luminous flux and radiant flux have same dimensions

  2. Luminous flux and luminous intensity have same dimensions

  3. Radiant flux and power have same dimension

  4. Relative luminosity is a dimensionless quantity

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Correct Option: A
Explanation:

Luminous flux ( Luminous power) is the amount of perceived power of light. It is measure energy.

Radiant flux (Radiant power) is the amount of radiated power.
While luminous intensity is ratio of luminous flux to solid angle.

A current carrying wire produces a magnetic field in its surrounding space.
The S.I. unit of magnetic flux density is 

physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

  1. Henry

  2. Tesla

  3. $AM^2$

  4. A-m

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Correct Option: B
Explanation:

The SI unit of Magnetic Flux density is $ Weber/m^2 $ which is also called Tesla.

The dimensional formula of magnetic flux is ___________.

physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

  1. $\left[ M{ L }^{ 2 }{ T }^{ -2 }{ A }^{ -1 } \right] $

  2. $\left[ M{ L }^{ 2 }{ T }^{ -3 }{ A }^{ -1 } \right] $

  3. $\left[ { M }^{ -1 }{ L }^{ -2 }{ T }^{ 2 }{ A }^{ 1 } \right] $

  4. $\left[ M{ L }^{ 3 }{ T }^{ -2 }{ A }^{ -1 } \right] $

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Correct Option: A
Explanation:

Dimensions of ${ \phi  } _{ B }=$ Unit of $B\times $ Unit of $A$
$=\dfrac { newton }{ ampere-metre } \times { metre }^{ 2 }=\dfrac { newton\times metre }{ ampere } $
$=\dfrac { kg{ ms }^{ -2 }\times m }{ A } =kg{ m }^{ 2 }{ s }^{ -2 }{ A }^{ -1 }=\left[ M{ L }^{ 2 }{ T }^{ -2 }{ A }^{ -1 } \right] $

In an experiment to measure the velocity of electrons, an electric field $(E)$ and a magnetic field $(B)$ are employed to produce zero deflection. Then :

physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

  1. The two fields are parallel and the velocity is given by $BE$

  2. The two fields are perpendicular and the velocity is given by $E/B$

  3. The two fields are parallel and the velocity is given by $E/B$

  4. The two fields are perpendicular and the velocity is given by $BE$

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Correct Option: B
Explanation:

The electric force on electron of charge ${e}$  is,
             $  F= {e}E $ 
                    where, E = electric field.

And Magnetic force ,
            $ F= {e}$ $(V\times B)$
               where, V= velocity of electron in magnetic field
                            B= magnetic field

From the above equations,

          $E= V\times B$
Electric field E is perpendicular to both velocity of electron and the magnetic field. So,  The two field is perpendicular to each other and the velocity of electron is $\dfrac{E}{B}$.

B. The two fields are perpendicular and the velocity is given by $\dfrac{E}{B}$.

Write the dimensions of Magnetic flux in terms of mass, time, length and charge.

physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

  1. $[M^1L^2T^{-1}Q^{-2}]$

  2. $[M^1L^2T^{-1}Q^{-1}]$

  3. $[M^1L^3T^{-1}Q^{-1}]$

  4. $[M^4L^2T^{-1}Q^{-1}]$

Show answer
Correct Option: B
Explanation:

B. $[M^1 L^2 T^ {-1}Q^{-1}]$


Dimension of magnetic field, 


$[B]=[M^1L^0Q^{-1}]$ 

Dimension of surface Area,

 $[A]=[L^2]$

The magnetic flux , 

$\phi=B.A$

The dimension of magnetic flux in term of mass, time, length and charge,
$[\phi]= [M^1L^2T^{-1}Q^{-1}]$