Tag: magnetic field due to bar magnet

Questions Related to magnetic field due to bar magnet

When a bar magnet is placed perpendicular to a uniform a magnetic field, it is acted upon by a couple of magnitude $1.732\times 10^{-5}Nm$. The angle through which the magnet should be turned so that the couple acting on it becomes $1.5\times 10^{-5}Nm$ is

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

  1. $60^{o}$

  2. $45^{o}$

  3. $30^{o}$

  4. $75^{o}$

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Correct Option: C
Explanation:

$\theta _{1}=\pi /2$
$z _{1}=1.732\times 10^{-5}$
$z _{2}=1.5\times 10^{-5}$
$z=mB\sin\theta $
$z _{max}=mB$
$=1.732\times 10^{-5}$
$z _{2}=mB\sin\theta ^{1}$
$1.5\times 10^{-5}=1.732\times 10^{-5}\sin\theta ^{1}$
$\sin\theta ^{1}=\dfrac{\sqrt{3}}{2}$
$\theta ^{1}=60$
Angle turned$=\pi /2-\theta ^{1}$
$=30^o$

A torque of 25 N m acts on a current carrying coil of area ${  5 m }^{ 2 }$ in a magnetic field of induction ${ 2  Wb/m }^{ 2 }$ . The angle between normal to coil and magnetic induction is ${ 30 }^{ \circ  }$ .Then value of current is 

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

  1. $0.4 A$

  2. $0.5 A$

  3. $400 mA$

  4. $5 A$

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Correct Option: D

A bar magnet of length 0.2 m and pole strength 5 A.m. should be kept in a uniform magnetic field of induction 15 tesla at angle ..... radians to the field so that the torque experienced by it will be 7.5N-m

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

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

  2. $\dfrac{\pi }{3}$

  3. $\dfrac{\pi }{6}$

  4. $\dfrac{\pi }{4}$

Show answer
Correct Option: C
Explanation:

$l=0.2$
$P=50m$
$B=15T$
$z=7.5Nm$
$\vec{z}=\vec{m}\times \vec{B}   m=lP$
$z=mBsin\theta $
$7.5=0.2\times 5\times 15\times sin\theta $
$sin\theta =\dfrac{1}{2}$
$\theta =\pi /6$

A bar magnet of magnetic moment $\overrightarrow{M}$, is placed in magnetic field of induction $\overrightarrow{B}$. The torque exerted on it is

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

  1. $\overrightarrow{ M }\cdot \overrightarrow{ B } $

  2. $-\overrightarrow{ M }\cdot \overrightarrow{ B } $

  3. $\overrightarrow{ M } \times \overrightarrow{ B } $

  4. $-\overrightarrow{ B } \times \overrightarrow{ M } $

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Correct Option: C
Explanation:

Torque is vector product of magnetic moment and the magnetic field.

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

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

  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 } } $

Show answer
Correct Option: B
Explanation:

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


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

A magnetic needle suspended freely orients itself

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

  1. in a definite direction

  2. in no direction

  3. upward

  4. downward

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Correct Option: A
Explanation:

A magnetic needle suspended freely orientates itself in a definite direction.

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 

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

  1. $175 \hat{k} N-m$

  2. $75 \hat{k} N-m$

  3. $150 \hat{k} N-m$

  4. $25 \sqrt{37}\hat{k} N-m$

Show answer
Correct Option: C
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, $\tau =\,150\hat{k}\,\,N-m$ 

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 _____.

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

  1. $90^0$

  2. $60^0$

  3. $45^0$

  4. $30^0$

Show answer
Correct Option: D
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
${\tau/2}{\tau}=\dfrac{MBsinT}{MB}$
$sinT=\dfrac{1}{2}$
$T=30^o$

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 :-

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

  1. $\hat{i}-\hat{j}-2\hat{k}$

  2. $\hat{i}+\hat{j}+2\hat{k}$

  3. $Zero$

  4. $None\ of\ these$

Show answer
Correct Option: A
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 $\hat{i}-\hat{j}-2\hat{k}$

 

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_

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

  1. $zero$

  2. $\pi irB$

  3. $2 \pi irB$

  4. $irB$

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Correct Option: A
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.