Tag: the bar magnet

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Questions Related to the bar magnet

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 closely wound solenoid of $2000$ turns and area of cross-section $1.5\times10^{-4}\ m^{2}$ carries a current of $2.0A$. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5\times10^{-2}$ Tesla making an angle of $30^{o}$ with the axis of the solenoid. The torque on the solenoid will be-

  1. $1.5\times10^{-3}\ N.m$

  2. $1.5\times10^{-2}\ N.m$

  3. $3\times10^{-2}\ N.m$

  4. $3\times10^{-3}\ N.m$
Reveal answer Fill a bubble to check yourself
D Correct answer
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

Two bar magnet are kept together and suspended freely in earth's magnetic field. When both like poles are aligned, the time period is 6 sec. When opposite poles are aligned, the time period is 12 sec. The ratio of magnetic moments of the two magnets is :

  1. $5/3$

  2. $2/1$

  3. $3/2$

  4. $3/1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$T _1= 6 = 2\pi \sqrt{\dfrac{1}{(M _1+M _2)B}}$
$T _2= 12 = 2\pi \sqrt{\dfrac{1}{(M _1+M _2)B}}$
$\therefore \dfrac{6}{12}=\dfrac{1}{2}= \sqrt{\dfrac{M- M _2}{M _1+M _2}}$ or $ M _1+M _2= 4( M _1-M _2) $
$ 3M _1= 5M _2$
$\dfrac{M _1}{M _2}=\dfrac{5}{3}$

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

Puneet peddles a stationary bicycle. The peddles are attached to a $100$ turn coil of area $0.10 \ m^2$. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of $0.01 \ T$ perpendicular to the axis of rotation of the coil. What is maximum voltage generated in the coil? 

  1. $0.314 \ V$

  2. $0.615 \ V$

  3. $0.921 \ V$

  4. $0.084 \ V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Maximum voltage (EMF) is given by E_max = N * B * A * omega. Frequency f = 0.5 rev/s, so omega = 2 * pi * f = 2 * pi * 0.5 = pi rad/s. E_max = 100 * 0.01 * 0.10 * pi = 0.1 * pi = 0.314 V.

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 length $16 cm$ has a pole strength of 500 milli amp.m. The angle at which it should be placed to the direction of external magnetic field of induction $2.5$ gauss so that it may experience a torque of $\sqrt { 3 } \times{ 10 }^{ -5 }$ N.m. is

  1. $\pi $

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

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

  4. $\dfrac { \pi }{ 6 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
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 magnetic moment $M$ and moment of inertia I is freely suspended such that the magnetic axial line is in the direction of magnetic meridian. If the magnet is displaced by a very small angle $(\theta )$ , the angular acceleration is (Magnetic induction of earths horizontal field $ = $ $B _H$)

  1. $\dfrac{MB _{H}\theta }{I}$

  2. $\dfrac{IB _{H}\theta }{M}$

  3. $\dfrac{M\theta }{IB _{H}}$

  4. $\dfrac{I\theta }{MB _{H}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Torque acting on the magnet is  $T=MB _H sin \theta $
Since $ \theta$ is very small , $T = MB _H \theta $
$I {\alpha} =  MB _H \theta $
$ \alpha = \dfrac{MB _H \theta}{I} $

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

When a bar magnet is suspended freely in a uniform magnetic field, identify the correct statements:

a) The magnet experiences only couple and undergoes only rotatory motion

b) The direction of torque is along the suspension wire

c) The magnitude of torque is maximum when the magnet is normal to the field direction

  1. only a and c are correct

  2. only a and b are correct

  3. only b and c are correct

  4. a, b, c are correct
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The bar magnet will only
experience a torque, It will not
experience any force
The direction of the torque will be
along the suspended wire according to
right hand thumbrule.
$\vec{z}=\vec{m}\times \vec{B}$
$=mB sin \theta .$
And its magnitude will be maximum if $\theta =90$


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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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. $\theta=45^o$

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

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 

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

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

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

  4. $\sqrt { 8 } B$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Magnetic field at the center B = mu0 * I / (2 * R). Magnetic field at distance x along the axis is B_axis = mu0 * I * R^2 / (2 * (R^2 + x^2)^(3/2)). For x = R, B_axis = mu0 * I * R^2 / (2 * (2 * R^2)^(3/2)) = mu0 * I * R^2 / (2 * 2 * sqrt(2) * R^3) = mu0 * I / (4 * sqrt(2) * R) = B / (2 * sqrt(2)) = B / sqrt(8).

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 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 :

  1. $\pi $

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

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

  4. $\dfrac{\pi }{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $z =\sqrt{3}\times 10^{-5}N.m$

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