Tag: magnetic effects of current and magnetism

Quizzes (16) Questions (174)

Questions Related to magnetic effects of current and magnetism

Multiple choice electrostatic and magnetic analogy magnetism and matter magnetic effects of current and magnetism physics

Force between two identical bar magnets whose centres are  $r meters $ apart is $4.8 N,$ when their axis are in the same line. If the separation is increased to $2r$ meters, the force between them is reduced to:

  1. 2.4 N

  2. 1.2 N

  3. 0.6 N

  4. 0.3 N
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$F=\dfrac{\mu _{0}m _{1}m _{2}}{4\pi d^{2}}$
$F\propto \dfrac{1}{d^{2}}$
$Fd^{2}=Constant$
$F _{1}d _{1}^{2}=F _{2}d _{2}^{2}$
$4.8\times r^2=F _{2}(2r)^2$
$F _{2}=1.2  N$

Multiple choice electrostatic and magnetic analogy magnetism and matter magnetic effects of current and magnetism physics

The magnetic induction at a distance d from the magnetic pole of the unknown strength m is B. If an identical pole is now placed at a distance of 2d from the first pole, the force between the two poles is          

  1. mB
  2. $\frac{mB}{2}$
  3. $\frac{mB}{4}$

  4. 2mB
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$B=\dfrac{\mu _{0}m}{4\pi d^{2}}$
$r=2d$
$F=\dfrac{\mu _{0}m _{1}m _{2}}{4\pi r^{2}}$
$F=\dfrac{\mu _{0}\times m\times m}{4\pi (d)^{2}}$
$F=\dfrac{\mu _{0}m^{2}}{4\pi \times 4d^{2}}$
$F=\dfrac{mB}{4}N$

Multiple choice electrostatic and magnetic analogy magnetism and matter magnetic effects of current and magnetism physics

Two magnetic poles have their strengths in the ratio 3 : 2. They are kept at a distance of 0.6 m in air and the force of repulsion is found to be 0.06 dynes. The pole strengths are (in amp. m)

  1. 1.8, 1.2

  2. 18, 12

  3. 6, 4

  4. 0.6, 3.6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$1 dyne = 10^{-5}$
$\mu _{0}=4\pi \times 10^{-7}TmA^{-1}$ 
$m _{1}=3m$
$m _{2}=3m$
$F=\dfrac{\mu _{0}m _{1}\times m _{2}}{4\pi \times r^{2}}$ r=0.6m
$0.06\times 10^{-5}=\dfrac{4\pi \times 10^{-7}\times 3m\times 2m}{4\pi \times (0.6)^{2}}$
$m^{2}=(0.6)^{2}$
$m=0.6$
$m _{1}=1.8$
$m _{2}=1.2$




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 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
B Correct answer
Explanation

The magnetic moment of the solenoid is M = N * I * A = 2000 * 2.0 * 1.5 * 10^-4 = 0.6 A*m^2. The torque in a uniform magnetic field B is given by tau = M * B * sin(theta), where theta is the angle between the magnetic field and the axis of the solenoid. Thus, tau = 0.6 * (5 * 10^-2) * sin(30 deg) = 0.6 * 0.05 * 0.5 = 1.5 * 10^-2 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

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
C Correct answer
Explanation

The magnetic dipole moment of the bar magnet is M = pole strength * length = (500 * 10^-3 A*m) * (0.16 m) = 0.08 A*m^2. The external magnetic field B = 2.5 gauss = 2.5 * 10^-4 Tesla. Using the torque formula tau = M * B * sin(theta), we have sqrt(3) * 10^-5 = 0.08 * (2.5 * 10^-4) * sin(theta) = 2 * 10^-5 * sin(theta). Solving for sin(theta) gives sin(theta) = sqrt(3) / 2, which means theta = 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

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 $\theta =90$