Tag: force on current carrying conductor

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Multiple choice 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

At a place the horizontal component of earth's field is $0.5 \times 10^{-4}T$. A bar magnet suspended horizontally perpendicular to earth's field experiences a torque of $4.5\times 10^{-4}N-m$ at that place. The magnetic moment of the magnet is: 

  1. $2.25\times 10^{-8}J/T$

  2. $1/9 J/T$

  3. $2.25 J/T$

  4. $9 J/T$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\tau = MB sin\theta$
$4.5 \times 10^{-4}= M\times 0.5 \times 10^{-4}. sin 90^0$
$M= \dfrac{4.5 \times 10^{-4}}{0.5\times 10^{-4}}= 9 J/T$

Multiple choice 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

A coil in the shape of equilateral tringale of side $0.02\ m$  is suspended from the vertex such that it is hanging in a  vertical place between the pole-pieces of a permanent magnet  producing a horizontal magnetic field of $5 \times 10^{-2}\  T.$ When a current of $0.1\ A$ passed through it and the  magnetic field is parallel to its plane then couple acting on  the coil is :

  1. $8.65 \times 10^{-7}\ N-m$

  2. $6.65 \times 10^{-7}\ N-m$

  3. $3.35 \times 10^{-7}\ N-m$

  4. $3.91 \times 10^{-7}\ N-m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The torque acting on a coil is given by,

$\tau $ = N B I A sin$\theta $
Here, A = $\frac{1}{2}  \times  base  \times $ height
$\begin{array}{l} =\frac { 1 }{ 2 } \times 0.02\times \sqrt { { { 0.02 }^{ 2 } }-{ { 0.01 }^{ 2 } } }  \ =1.732\times { 10^{ -4 } }{ m^{ 2 } } \end{array}$
$\theta $ = Angle between direction of the magnetic field and normal to the plane of the coil. Also, as the magnetic field is parallel to the plane of the coil, so $\theta $ = 90
thus, 
$\tau$  = $1\times 5\times { 10^{ -2 } }\times 0.1\times 1.732\times { 10^{ -4 } }\times 1$
    = 8.6$\times10^{-7}$ Nm



Multiple choice 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

A bar magnet when placed at an angle of $30^o$ to the direction of magnetic field of induction of $ 5\times10^{-5} T$, experiences a moment of a couple $2.5\times10^{-6} N-m$. If the length of the magnet is $5\ cm$ its pole strength is:

  1. $2\times10^{-2}\ Am$

  2. $5\times10^{-2}\ Am$

  3. $2\ Am$

  4. $5\ Am$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Torque experienced by bar magnet is $\tau =MB\sin \theta $ where $M$magnetic moment, $B$ is magnetic field induction and $\theta $ is the angle between $M\,\,and\,\,B$

It is given that

  $ \tau =2.5\times {{10}^{-6}}N $

 $ B=5\times {{10}^{-5}}T $

 $ L=0.05m $

 $ 2.5\times {{10}^{-6}}=M\times 5\times {{10}^{-5}}\,\times \dfrac{1}{2} $

 $ M={{10}^{-1}} $

 $ M=m\times L $

 $ {{10}^{-1}}=m\times 0.05 $

 $ m=2 Am $

Multiple choice 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

A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it, then the angle by which it is to be rotated is:

  1. $30^o$

  2. $45^o$

  3. $60^o$

  4. $90^o$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that,

$ \tau =MB\sin \theta  $

$ \tau \propto \sin \theta  $

So,

$ \dfrac{{{\tau } _{1}}}{{{\tau } _{2}}}=\dfrac{MB\sin {{\theta } _{1}}}{MB\sin {{\theta } _{2}}} $

$ \dfrac{{{\tau } _{1}}}{{{\tau } _{2}}}=\dfrac{\sin {{\theta } _{1}}}{\sin {{\theta } _{2}}} $

$ \dfrac{\tau }{\frac{\tau }{2}}=\dfrac{\sin {{90}^{0}}}{\sin {{\theta } _{2}}} $

$ \sin {{\theta } _{2}}=\dfrac{1}{2} $

$ {{\theta } _{2}}={{30}^{0}} $

Hence, the angle is ${{30}^{0}}$


Multiple choice 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

If a current is passed through a loop which is placed in a magnetic field, then the acting torque will not depend on

  1. Shape of the loop

  2. Area of the loop

  3. The current value

  4. Magnetic field
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since,  $T _B=MB\sin\theta$ where $\theta$ is angle between body and the magnetic field$\implies $ shape of loop does not affect torque since $B$ depends on the area and current.

Multiple choice 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

A charged particle is moving with uniform velocity $V\hat {j}$ through a uniform magnetic field $B(-\hat {i})$ and a unifom electric field $\vec {E}$. Then $\vec {E}$ is

  1. $-Bv\hat {k}$

  2. $Bv\hat {k}$

  3. $\dfrac {v}{B}\hat {k}$

  4. $\dfrac {-B}{v}\hat {k}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice 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

At some location on earth the horizontal components of earth's magnetic field is $18 \times 10^-6 T.$ At this location, magnetic needle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes $45^0$ angle with horizontal in equilibrium to keep this needle horizontal, the vertical force that should be applied at one of its ends is:

  1. $3.6 \times 10^-5 N$

  2. $6.5 \times 10^-5 N$

  3. $1.3 \times 10^-5 N$

  4. $1.8 \times 10^-5 N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At $45 ^o, B _H = B _V$

$F \dfrac{l}{2} = MB _V = m \times l \times B _V$
$F = \dfrac{2 m l B _V} { l} = 3.6 \times 18 \times 10^{-6} $
$ = 6.5 \times 10^{-5} N$

Multiple choice 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

Two bar magnets with magnetic moments2$\mathrm { M }$ and $\mathrm { M }$ are fastened together at right angles to each other at their centres to forma crossed system, which can rotate freelyabout a vertical axis through the centre. The crossed system sets in earth's magnetic fieldmaking an angle $\theta$ with the magnetic merid-ian such that

  1. $\theta = \tan ^ { - 1 } \left( \dfrac { 1 } { \sqrt { 3 } } \right)$

  2. $\theta = \tan ^ { - 1 } ( \sqrt { 3 } )$

  3. $\theta = \tan ^ { - 1 } \left( \dfrac { 1 } { 2 } \right)$

  4. $\theta = \tan ^ { - 1 } \left( \dfrac { 3 } { 4 } \right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The system is in equilibrium when the net torque from the two magnets balances the torque from the Earth's magnetic field. The magnetic moments are 2M and M at 90 degrees. The resultant magnetic moment is sqrt((2M)^2 + M^2) = M sqrt(5), and the angle alpha it makes with the 2M magnet is tan(alpha) = M/2M = 1/2. The equilibrium angle theta with the meridian satisfies tan(theta) = M_perp / M_parallel = 1/2.

Multiple choice 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

If the net force acting on the loop is zero then : 

  1. no torque acts on loop

  2. loop performs translational motion

  3. torque can act on the loop if lines of force do not coincide

  4. both a) and b)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If the net force acting on the loop is zero then, 


1) The loop performs rotational motion.


2) No torque acts on the loop.

The correct option is A.