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Questions Related to physics

A circular coil of $25$ turns and radius of $12$cm is placed in a uniform magnetic field of $0.5$ T normal to the plane of coil. If the current in the coil is $5$A, then total torque experienced by 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. $1.5$N m

  2. $2.5$N m

  3. $3.5$ N m

  4. zero

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Correct Option: D
Explanation:

Here, n = $25$ turns, r = $12$cm, B = $0.5$T
Since the coil is placed in uniform magnetic field normal to the place of the coil. Hence the angle between magnetic moment and magnetic field direction is zero $(i.e. \theta = 0)$
$therefore = mB sin \theta = mB sin 0$
$\therefore T = 0$

A closely wound solenoid of 2000 turns and area of cross-section $1.5 \times 10^{-4}m^2$ carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{-2}T$, making an angle of 30$^o$ with the axis of the solenoid. The torque on the solenoid 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. $3 \times 10^{-3} N -m$

  2. $1.5 \times 10^{-3} N -m$

  3. $1.5 \times 10^{-2} N -m$

  4. $3 \times 10^{-2} N -m$

Show answer
Correct Option: C
Explanation:

Given :   $N = 2000$ turns        $A = 1.5\times 10^{-4} m^2$              $I = 2.0$ A                $B = 5\times 10^{-2}$ T            $\theta = 30^o$

Torque    $\tau = NI (A\times B) =NIAB \sin \theta$                 
$\therefore$  $\tau = (2000) (2.0) (1.5\times 10^{-4}) (5\times 10^{-2}) (0.5) $                           $(\because \sin 30^o =0.5)$
$\implies$   $\tau = 1.5\times 10^{-2}$  $N-m$

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: 

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

  2. $1/9 J/T$

  3. $2.25 J/T$

  4. $9 J/T$

Show answer
Correct Option: D
Explanation:

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

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 :

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

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

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

  4. $3.91 \times 10^{-7}\ N-m$

Show answer
Correct Option: A
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



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:

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

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

  3. $2\ Am$

  4. $5\ Am$

Show answer
Correct Option: C
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 $

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:

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. $30^o$

  2. $45^o$

  3. $60^o$

  4. $90^o$

Show answer
Correct Option: A
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 ${{30}^{0}}$


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

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. Shape of the loop

  2. Area of the loop

  3. The current value

  4. Magnetic field

Show answer
Correct Option: A
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.

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

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. $-Bv\hat {k}$

  2. $Bv\hat {k}$

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

  4. $\dfrac {-B}{v}\hat {k}$

Show answer
Correct Option: B

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:

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. $3.6 \times 10^-5 N$

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

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

  4. $1.8 \times 10^-5 N$

Show answer
Correct Option: B
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$

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

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

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

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

  4. $\theta = \tan ^ { - 1 } \left( \dfrac { 3 } { 4 } \right)$

Show answer
Correct Option: C