Tag: magnetic effects of current and magnetism

Quizzes (16) Questions (174)

Questions Related to magnetic effects of current and magnetism

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 $z =\sqrt{3}\times 10^{-5}N.m$

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

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 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

  1. $60^{o}$
  2. $45^{o}$
  3. $30^{o}$
  4. $75^{o}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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$=\pi /2-\theta ^{1}$
$=30^o$

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 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 

  1. $0.4 A$
  2. $0.5 A$
  3. $400 mA$
  4. $5 A$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Torque = N * I * A * B * sin(theta). Here N=1 (implied). Torque = 25, A = 5, B = 2, theta = 30 degrees. 25 = I * 5 * 2 * sin(30) = I * 10 * 0.5 = 5 * I. I = 25 / 5 = 5 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

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

  1. $\dfrac{\pi }{2}$
  2. $\dfrac{\pi }{3}$
  3. $\dfrac{\pi }{6}$
  4. $\dfrac{\pi }{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $
$sin\theta =\dfrac{1}{2}$
$\theta =\pi /6$

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

  1. $\overrightarrow{ M }\cdot \overrightarrow{ B } $
  2. $-\overrightarrow{ M }\cdot \overrightarrow{ B } $
  3. $\overrightarrow{ M } \times \overrightarrow{ B } $
  4. $-\overrightarrow{ B } \times \overrightarrow{ M } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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