Tag: magnetic effects of current and magnetism

Questions Related to magnetic effects of current and magnetism

An electric dipole is placed at an angle of ${30}^{o}$ with an electric field of intensity $2 \times { 10 }^{ 5 }N\quad { C }^{ -1 }$. It experiences a torque equal to $4 \ N$ $m$. The charge on the dipole of the dipole length is $2 \ cm$ is

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. $8 \ mC$

  2. $4 \ mC$

  3. $2 \ mC$

  4. $6 \ mC$

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Correct Option: D
Explanation:

Here $E=2\times { 10 }^{ 5 }N{ C }^{ -1 },l=2cm,\tau =4Nm\quad $
Torque $\vec { \tau  } =\vec { p } \times \vec { E } =pE\sin { \theta  } $
$\therefore 4=p\times 2\times { 10 }^{ 5 }\times \sin { { 30 }^{ o } } $
or $p=4\times { 10 }^{ -5 }Cm$
$\therefore$ Charge $q=\cfrac { p }{ l } =\cfrac { 4\times { 10 }^{ -5 }Cm }{ 0.02m } =2\times { 10 }^{ -3 }C=2mC$

The potential energy of a bar magnet of magnetic moment $M$ placed in a magnetic field of induction $B$. The position of stable equilibrium of the magnet is at the angular position given by $\theta$ equal to:
[where $\theta$ is the angle between B and M]
force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. ${0}^{o}$

  2. ${90}^{o}$

  3. ${45}^{o}$

  4. ${180}^{o}$

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Correct Option: A

A circular coil of 25 turns and radius 12 cm is placed in a uniform magnetic field of 0.5 T normal to the plane of the coil. If the current in the coil is 6 A then total torque acting on the coil is

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. zero

  2. 3.4 N m

  3. 3.8 N m

  4. 4.4 N m

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Correct Option: A
Explanation:

The torque acting on the coil $\tau$=|$\bar m$ X $\bar B$ | =$ mB\sin\Theta$
Here the circular coil is placed normal to the direction of magnetic field then the angle between the. direction of magnetic moment ($\bar m$) and magnetic field ($\bar B$) is zero, then
$\tau$ = $mBsin\Theta$ = $mBsin 0$ = 0                $\tau = 0$

The magnitude of torque experienced by a square coil of side 12 cm which consists of 25 turns and carries a current 10 A suspended vertically and the normal to the plane of coil makes an angle of 30 with the direction of a uniform horizontal magnetic field of magnitude 0.9 T is:

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. 1.6 Nm

  2. 1.2 Nm

  3. 1.4 Nm

  4. 1.8 Nm

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Correct Option: A
Explanation:

Here,
$N=25,\\ I=10A, \\B= 0.9T,\\\theta=30^0$

$A=a^2=12 \times 10^{-2} \times 12 \times 10^{-2} =144 \times 10^{-4}m^2$

       $\tau$= $NIA\sin\theta$

$\therefore \tau = 25 \times 10\times 144 \times 10^{-4} \times 0.9 \times \sin30^0 = 1.6Nm$  

A circular coil of 70 turns and radius 5 cm carrying a current of 8 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.5 T. The field lines make an angle of 30 with the normal of the coil then the magnitude of the counter torque that must be applied to prevent the coil from turning is :

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. 33 N m

  2. 3 N m

  3. 3.3 x $10^{2}$N m

  4. 3.3 x $10^{-4}$N m

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Correct Option: C
Explanation:
Number of turns on the circular coil, $n = 70$

Radius of the coil, $r = 5.0 cm = 0.05\ m$

Area of the coil , $A= \pi r^2=3.14\times(0.05)^2= $

Current flowing in the coil, $I = 8.0\ A$

Magnetic field strength, $B = 1.5\ T$

Angle between the field lines and normal with the coil surface,

$θ = 30^o$

The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the relation,

$\tau = n IBA sinθ$ … (i)

$\tau= 70 × 8 × 1.5 × 0.00785× sin30^o$

$\tau = 3.3\ N m$


A dipole of magnetic moment $\vec{m}=30\hat{j}$A $m^2$ is placed along the y-axis in a uniform magnetic field $\bar{B}=(2\hat{i}+5\hat{j})$T. The torque acting on it is?

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. $-40\hat{k}$N m

  2. $-50\hat{k}$N m

  3. $-60\hat{k}$N m

  4. $-70\hat{k}$N m

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Correct Option: C
Explanation:

Given that,

Magnetic moment of a dipole, $\vec m=30\hat j\ Am^2$
Magnetic field, $\vec B= (2\hat i+ 5\hat j)\ T$

Torque acting on the dipole , $\vec{\tau}= \vec m \times \vec B$

$\implies  \vec{\tau}= 30\hat j \times( 2\hat i +5\hat j)= -60 \hat k \ Nm $

A circular coil of $100$ turns, radius $10$cm carries a current of $5$A. It is suspended vertically in a uniform horizontal magnetic field of $0.5$T and the field lines make an angle of $60^0$ with the plane of the coil. The magnitude of the torque that must be applied on it to prevent it from turning is going to be

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. $2.93$ N m

  2. $3.43$ N m

  3. $3.93$ N m

  4. $4.93$ N m

Show answer
Correct Option: C
Explanation:

Given that,


Number of turns, $N=100$
Radius of coil, $r=10\ cm=0.1 \ m$
Current in the coil, $i=5\ A$
Magnetic moment of the coil, $M= Ni(\pi r^2)$
Magnetic field, $B=0.5\ T$
$\theta=90^o-60^o=30^o$

Torque that must be applied to the coil, $\tau = Ni(\pi r^2) \times B \times sin\ \theta=100\times 5\times (3.14\times 0.1^2)\times0.5\times sin 30^o=3.93\ Nm$

So, correct option is $(C)$

The torque and magnetic potential energy of a magnetic dipole in most stable position in a uniform magnetic field $(\bar{B})$ having magnetic moment $(\bar{m})$ will be

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. $-mB, zero$

  2. $mB, zero$

  3. $zero, mB$

  4. $zero, -mB$

Show answer
Correct Option: D
Explanation:

Torque, $\bar{T} = \bar{m} \times \bar{B} = mB sin \theta$
and magnetic potential energy $U _m = \bar{m}. \bar{B} = mB cos \theta$ at $\theta = 0^0$ the dipole will be in most stable position
$T = mB sin\theta = mB sin 0^0 = 0$
and $U _m = -mB cos\theta = -mB cos 0^0 = -mB$

The final torque on a coil having magnetic moment 25 A $m^2$ in a 5 T uniform external magnetic field, if the coil rotates through an angle of 60 under the influence of the magnetic field is:

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. 216.5 Nm

  2. 108.25 Nm

  3. 102.5 Nm

  4. 58.1 Nm

Show answer
Correct Option: B
Explanation:
Here, 
$m=25 A m^2 ; \\ \theta  = 60^0 ;\\ B = 5T$


$|\vec {\tau} = |\vec m \times \vec B |=mB\sin\Theta$

$\therefore  = 25 \times 5\times  \sin60^0$ or

 $\tau= 125 \times \dfrac{\sqrt3}{2} = 108.25 Nm$

The magnetic moment of a short bar magnet placed with its magnetic axis at $30^0$ to an external field of $900$G and experiences a torque of $0.02$N m is going to be

force and torque on a current carrying rectangular loop in a uniform magnetic field torque on current carrying loop force on current carrying conductor magnetic effects of current and magnetism physics

  1. $0.35 A m^2$

  2. $0.44 A m^2$

  3. $2.45 A m^2$

  4. $1.5 A m^2$

Show answer
Correct Option: B
Explanation:

Here, $B = 900 Gauss = 900 \times 10^{-4}T. = 9 \times 10^{-2}T$
$T = 0.02 N m and \theta = 30^0$
$\therefore T = mB sin \theta$
$\Rightarrow 0.02 = m \times 9 \times 10^{-2} \times sin 30^0$
$m = \dfrac{0.02 \times 2 }{9 \times 10^{-2}}= 0.44 A m^2$.