Tag: magnetic effects of current and magnetism

Questions Related to magnetic effects of current and magnetism

A magnetic needle is kept in a non uniform magnetic field. It experiences :

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. a force and a torque

  2. a force but not a torque

  3. torque but not a force

  4. neither a torque nor a force

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Correct Option: A
Explanation:

Non uniform magnetic field gives rise to different forces at the different points on the needle thus producing a net force . This also gives rise to net torque

When a bar magnet is suspended in a uniform magnetic field, the torque acting on it will be :

a) maximum e) $\theta=45^o$ with the field
b) half the maximum field f) $\theta=60^o$ with the field
c) $\sqrt{3}/2$ times the maximum field g) $\theta=30^o$ with the field
d) $1/\sqrt{2}$ times the maximum field h) $\theta=90^o$ with the field
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. a-g, b-h, c-d, d-e

  2. a-e, b-f, c-g, d-h

  3. a-f, b-e, c-g, d-h

  4. a-h, b-g, c-f, d-e

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Correct Option: D
Explanation:

Torque acting on the bar magnet suspended in a uniform field B is given by,

$T=MBsin\theta$  where $M$  is the moment of magnet.
Torque is maximum when $sin\theta=1$ i.e. $\theta=90^o$
$\therefore$ maximum torque $=MB$
Torque is half when $sin\theta=1/2$ i.e. $\theta=30^o$
Torque is $\sqrt{3}/2$ the maximum when $sin\theta=\sqrt{3}/2$ i.e. $\theta=60^o$
Torque is $1/\sqrt{2}$ the maximum when $sin\theta=1/\sqrt{2}$ i.e. $\theta=45^o$

Magnetic field at the center of a circular coil of radius R due to current I flowing through it is B The magnetic field at a point along the axis at distance R from the center 

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. $\dfrac { B } { 2 }$

  2. $\dfrac { B } { 4 }$

  3. $\dfrac { B }{ \sqrt { 8 } } $

  4. $\sqrt { 8 } B$

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Correct Option: C

A bar magnet of length $16 cm$ has a pole strength of $500\times 10^{-3}Am$ The angle at which it should be placed to the direction of external magnetic field of induction $2.5 G$ so that it may experience a torque of $\sqrt{3}\times 10^{-5}$ 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. $\pi $

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

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

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

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Correct Option: C
Explanation:

$l=16m$
$P=500\times 10^{-3}Am$
$z=\vec{m}\times \vec{B}$

$B=2.5\;G=2.5\times 10^{-4}T$
$m=lP  $
torque due to magnetic field $z =\sqrt{3}\times 10^{-5}N.m$

$z=mBm \times sin Q$
$\sqrt{3}\times 10^{-5}=lPBsinQ$
$\sqrt{3}\times 16\times 500\times 2.5\times sinQ.$
$SinQ=\dfrac{\sqrt{3}}{2}$
$Q=\pi /3$

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. $75^{o}$

Show answer
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 } } $

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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.