Tag: electromagnetic induction

Questions Related to electromagnetic induction

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

Show answer
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] $

Show answer
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 $BE$

Show answer
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,
$[\phi]= [M^1L^2T^{-1}Q^{-1}]$

A solenoid coil is wound on a frame of rectangular cross section. If all the linear dimension of the frame are increased by a factor of two and the number of turns per unit length of the coil remains the same, the self-inductance increases by a factor of

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $4$

  2. $8$

  3. $12$

  4. $16$

Show answer
Correct Option: B
Explanation:

Self inductance, $L = \dfrac{\mu 0 N^2 A}{l}$ _(1)


where N = total number of turns


$n = \dfrac{N}{l}$, where n = number of turns per unit length

$N = nl$ put it in (1)

$L = \dfrac{\mu _0 (n l )^2 A}{l} = \mu _0 n^2 l A$

In the given question n is same
Area will increase by $4$ times
length will increase by $2$ times

$L' = \mu _0 n^2 (2l) (4A) = 8 \mu _0 n^2 lA$

$L' = 8L$ inductance will increased by $8$ times.

The physical quantity which is measured in the unit of Wb $A^{-1}$ is

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. self inductance

  2. mutual inductance

  3. magnetic flux

  4. both (a) and (b)

Show answer
Correct Option: D
Explanation:

both (a) and (b) 

self inductance and mutual inductance has same unit both measure magnetic flux per unit area.

The unit of inductance is equivalent to

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $\dfrac {volt\times ampere}{second}$

  2. $\dfrac { ampere}{volt\times second}$

  3. $\dfrac {volt}{Ampere\times second}$

  4. $\dfrac {volt\times second}{ampere}$

Show answer
Correct Option: D
Explanation:

$As \, \varepsilon \, = \,  L \, \dfrac{dI}{dt},$
$L \, = \, \varepsilon \dfrac{dt}{dI} \Rightarrow \, L \, = \, \dfrac{volt \times second}{ampere}$

What is the SI unit of self-inductance ?

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. Henry

  2. Tesla

  3. Weber

  4. Gauss

Show answer
Correct Option: A
Explanation:

 Henry (symbol H) is the SI derived unit of self-inductance. 

Two concentric co-planar circular loops of radii $r _1$ and $r _2$ carry currents of respectively $i _1$ and $i _2$ in opposite directions (one clockwise and the other anticlockwise.) The magnetic induction at the centre of loops is half that due to $i _1$ alone at the centre. If $r _2 = 2r _1$. the value of $i _2 / i _1$ is 

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. $2$

  2. $\dfrac{1}{2}$

  3. $\dfrac{1}{4}$

  4. $1$

Show answer
Correct Option: D
Explanation:

$\begin{array}{l} B=\frac { { { \mu _{ o } }{ i _{ 1 } } } }{ { 2{ r _{ 1 } } } } -\frac { { { \mu _{ o } }{ i _{ 2 } } } }{ { 2\left( { 2{ r _{ 1 } } } \right)  } }  \ =\frac { { { \mu _{ o } }{ i _{ 1 } } } }{ { 2{ r _{ 1 } } } } -\frac { { { \mu _{ o } }{ i _{ 2 } } } }{ { 4{ r _{ 2 } } } }  \ { B^{ ' } }=\frac { { { \mu _{ o } }{ i _{ 1 } } } }{ { 2{ r _{ 1 } } } }  \ \Rightarrow B=\frac { { { B^{ ' } } } }{ 2 }  \ \Rightarrow \frac { { { \mu _{ o } }{ i _{ 1 } } } }{ { 2{ r _{ 1 } } } } -\frac { { { \mu _{ o } }{ i _{ 2 } } } }{ { 4{ r _{ 1 } } } } =\frac { { { \mu _{ o } }{ i _{ 1 } } } }{ { 4{ r _{ 1 } } } }  \ \Rightarrow \frac { { { \mu _{ o } }{ i _{ 1 } } } }{ { 4{ r _{ 1 } } } } =\frac { { { \mu _{ o } }{ i _{ 2 } } } }{ { 4{ r _{ 1 } } } }  \ \Rightarrow \frac { { { i _{ 1 } } } }{ { { i _{ 2 } } } } =1 \ Hence, \ option\, \, D\, \, is\, \, correct\, \, answer. \end{array}$

When the number of turns in a solenoid is doubled without any change in the length of the solenoid, its self-inductance becomes :

introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

  1. Half

  2. Double

  3. Four times

  4. Eight times

Show answer
Correct Option: C
Explanation:

Self-inductance,
$\begin{array}{l}L \propto {N^2}\\therefore \dfrac{{{L _1}}}{{{L _2}}} = {\left( {\dfrac{{{N _1}}}{{{N _2}}}} \right)^2} = {(2)^2} = 4...........{ since,{N _2} = 2{N _1}} \\left[ {{L _2} = 4{L _1}} \right]\end{array}$