Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

Torques ${ \tau  } _{ 1 }$ and ${ \tau  } _{ 2 }$ are required for a magnetic needle to remain perpendicular to the magnetic fields at two different places. The magnetic fields at those places are ${B} _{1}$ and ${B} _{2}$ respectively; then ratio $\cfrac{{B} _{1}}{{B} _{2}}$ is

  1. $\cfrac { { \tau } _{ 2 } }{ { \tau } _{ 1 } } $
  2. $\cfrac { { \tau } _{1 } }{ { \tau } _{ 2 } } $
  3. $\cfrac { { \tau } _{ 1 }+{ \tau } _{ 2 } }{ { \tau } _{ 1 }-{ \tau } _{ 2 } } $
  4. $\cfrac { { \tau } _{ 1 }-{ \tau } _{ 2 } }{ { \tau } _{ 1 }+{ \tau } _{ 2 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\tau = \bar M \times \bar B.$ since perpendicular, $\theta = 90^{\circ}$


$\tau _1 : \tau _2 $ = $M.B _1.sin90^{\circ}$:$M.B _2.sin90^{\circ}$=$B _1:B _2$

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A magnet of magnetic moment $50\hat{i} A-m^2$ is placed along x- axis in a magnetic field$ \overrightarrow {B} =( 0.5 \hat{i} + 3.0 \hat{j})$ tesla. The torque acting on the magnet is 

  1. $175 \hat{k} N-m$
  2. $75 \hat{k} N-m$
  3. $150 \hat{k} N-m$
  4. $25 \sqrt{37}\hat{k} N-m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,

Magnetic field $\vec{B}\,=\,0.5\hat{i}\,+\,3.0\hat{j}\,\,T$

Magnetic moment $\mu =50\hat{i}\,\,A{{m}^{2}}$

Torque on magnet  $\tau $

$\tau =\mu \times \vec{B}\,\,=\,\,50\hat{i}\,\times \,(0.5\hat{i}\,+\,3.0\hat{j}\,)$

$\tau =\,150\hat{k}\,\,N-m$

Hence, Torque experiences by the magnet present in the magnetic field, $\tau =\,150\hat{k}\,\,N-m$ 

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A bar magnet of dipole moment M is initially parallel to a magnetic field of induction B. The angle through which it should be rotated so that the torque acting on it is half the maximum torque is _____.

  1. $90^0$
  2. $60^0$
  3. $45^0$
  4. $30^0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\tau=MXB=MBsinQ$

At maximum torque the Q is $90^o$ 
$\tau=MBsin(90^o)=MB$----------------1

At the half of maximum torque,
$\dfrac{\tau}{2}=MXB=MBsinT$-----------------2

1 to 2
${\tau/2}{\tau}=\dfrac{MBsinT}{MB}$
$sinT=\dfrac{1}{2}$
$T=30^o$

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A magnetic dipole of dipole moment $10(\hat{i}+\hat{j}+\hat{k})$ is placed in a magnetic field $0.6\hat{i}+0.4\hat{j}+0.5\hat{k}$, force acting on the dipole is :-

  1. $\hat{i}-\hat{j}-2\hat{k}$
  2. $\hat{i}+\hat{j}+2\hat{k}$
  3. $Zero$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

Dipole moment $m=10\left( \hat{i}+\hat{j}+\hat{k} \right)$

Magnetic field $B=0.6\hat{i}+0.4\hat{j}+0.5\hat{k}$

We know that,

The force is

  $ \tau =m\times B $

 $ \tau =\left( 10\hat{i}+10\hat{j}+10\hat{k} \right)\times \left( 0.6\hat{i}+0.4\hat{j}+0.5\hat{k} \right) $

 $ \tau =\hat{i}-\hat{j}-2\hat{k} $

Hence, the force acting on the dipole is $\hat{i}-\hat{j}-2\hat{k}$

 

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A current $i$ is flowing in a circular conductor of radius $r$, it is lying in a uniform magnetic field $B$ such that its plane is normal to $B$. The magnetic force acting on the loop will be_

  1. $zero$
  2. $\pi irB$
  3. $2 \pi irB$
  4. $irB$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Lets consider a small arc of length $dl$ of the circular loop
The magnetic force acting on the loop will be,
$\vec F=i(d\vec l\times \vec B)$ .. . .  ..  . . .. (1)
the force acting towards the center.
When the magnetic field $B$ is perpendicular to the $dl$, the angle between them is $90^0$.
$\theta=90^0$
From equation (1),
$\vec F=i|d\vec l||\vec B|sin\theta=idlB$
But the similar element of arc is opposite to the consider arc, so the current is also in the opposite direction and the force acting on the loop is now towards outward, both force cancel each other and the net force acting on the loop will be zero.
The correct option is A.
Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

If a bar magnet is kept perpendicular in a magnetic field of unit magnetic induction then its magnetic moment is equal to:

  1. magnetic flux

  2. torque

  3. pole strength

  4. magnetic flux density
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From the given question, 

$B=1T$
$\theta=90^0$
The torque acting on the bar magnet is given by
$\tau=MBsin\theta=M\times 1\times sin90^0$
$\tau=M$
The magnetic moment is equal to the torque.
The correct option is B.

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

The small magnets each of magnetic moment $10A-{ m }^{ 2 }$ are placed end on position 0.1m apart from their centres.The force acting between them is :

  1. $0.6\times { 10 }^{ 7 }N$
  2. $0.06\times { 10 }^{ 7 }N$
  3. $0.6N$
  4. $0.06N$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,


$M _1=M _2=10Am^2$


$r=0.1m$

The force acting between the magnet is given by,

$F=\dfrac{\mu _0}{4\pi }.\dfrac{6M _1M _2}{r^4}$

$F=10^{-7}\times \dfrac{6\times 10\times 10}{(0.1)^4}$

$F=0.6N$

The correct option is C.

Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A magnetic needle lying parallel to a magnetic field requires W unit of work to turn it through $60^0$. The torque needed to maintain the needle in this position will be 

  1. $\sqrt3W$

  2. W
  3. $(\sqrt3/2)W$

  4. 2W
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Work done to rotate the needle$=$Potential energy of needle$=$W
If M$=$Magnetic dipole moment of the needle and B$=$magnetic field then:
$W=MB(\cos\theta _1-\cos\theta _2)$ (where $\theta _1=0^o$ and $\theta _2=60^o$)
$\Rightarrow W=MB(1-1/2)=\dfrac{MB}{2}$ or $MB=2W\rightarrow (1)$
So restoring torque acting on the needle is given by:
$\tau =M\times B=MB\sin 60^o$(In magnitude)
$\Rightarrow \tau =\dfrac{\sqrt{3}}{2}MB$
$\Rightarrow \tau =\dfrac{\sqrt{3}}{2}2W=\sqrt{3}W$.
Multiple choice physics magnetic effects of current and magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

A bar magnet having centre O has a length of 4 cm. Point $P _1$ is in the broad side-on and $P _2$ is in the end side-on position with $OP _1=OP _2=10$ metres. The ratio of magnetic intensities H at $P _1$ and $P _2$ is

  1. $H _1:H _2=16:100$
  2. $H _1:H _2=1:2$
  3. $H _1:H _2=2:1$
  4. $H _1:H _2=100:16$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a short magnet, magnetic intensity H at broad side-on (equatorial) is H1 = M / (4 * pi * d^3) and at end side-on (axial) is H2 = 2 * M / (4 * pi * d^3). The ratio H1:H2 = 1:2.