Tag: physics

Questions Related to physics

A current carrying loop is free to turn in a uniform magnetic field. The loop will then come into equilibrium when it's plane is inclined at

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. $0^{0}$ to the direction of the field.

  2. $45^{0}$ to the direction of the field.

  3. $90^{0}$ to the direction of the field.

  4. $60^{0}$to the direction of the field.

Show answer
Correct Option: C
Explanation:

loop will come to equilibrium when torque is zero
So, $\vec{\tau} = 0 = \vec{M}\times \vec{B}$
                            $= MB sin \theta$
                            $= NiA Bsin \theta$
So, angle between area vector and field is zero than only $\theta$ is 0 so that $sin \theta = 0$.
So, for area vector along field direction plane is perpendicular to the field.

A coil of area $100\ cm^{2}$ having $500$ turns carries a current of $1\ m A$. It is suspended in a uniform magnetic field of induction $10^{-3}Wb /m^{2}$ . It's plane makes an angle of $60^{o}$ with the lines of induction. The 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. $250\times 10^{-8}Nm$

  2. $25\times 10^{-8}Nm$

  3. $2.5\times 10^{-8}Nm$

  4. $0.2\times 10^{-8}Nm$

Show answer
Correct Option: A
Explanation:

$J=\vec { M } \times \vec { B } $
$=ni\ \vec { A}\times   \vec { B } $
$=1\times { 10 }^{ -3 }\times 100\times { 10 }^{ -4 } \times  { 10 }^{ -3 }\times 500\times \sin { { 30 }^{ o  } } $
$=25\times { 10 }^{ -7 }Nm.$

Since. Plane makes an angle of $60^o$ with the lines of induction, it will makes an angle of $30^o$ with it's normal.

A circular coil of $1$ turn and area $0.01 m$ carries a current of $10 A$. It is placed in a uniform magnetic field of induction $0.1T$ such that the plane of the circle is perpendicular to the direction of the field, the 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. $0.1N m$

  2. $0.001Nm$

  3. $0.01N m$

  4. $0$

Show answer
Correct Option: D
Explanation:

Magnetic moment vector is parallel to that of field. Hence, Torque $=\vec { M } \times \vec { B } =Q\quad $

A current carrying coil tends to set itself____

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. parallel to an external magnetic field.

  2. parallel to its own magnetic field

  3. perpendicular to the external magnetic field.

  4. perpendicular to the geographic meridian

Show answer
Correct Option: C
Explanation:

When current carrying coil is placed in the magnetic field torque acts on the coil. The torque acting on the coil will be zero if the coil and the magnetic field are perpendicular. Thus it orients such that the plane of coil is perpendicular to the field direction.

A rectangular coil of wire of area $400 \ cm^{2}$ contains $500$ turns. It is placed in a magnetic field of induction $4\times 10^{-3}\ T$ and it makes an angle $60^o$ with the field. A current of $0.2 \ A$ is passed through it. The torque 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. $8\sqrt{3}\times 10^{-3}\ N-m$

  2. $8\times 10^{-3}\ N-m$

  3. $8\sqrt{3}\times 10^{-4}\ N-m$

  4. $8\times 10^{-4}\ N-m$

Show answer
Correct Option: B
Explanation:

$A=400\times { 10 }^{ -4 }{ m }^{ 2 }$
 $n = 500$
$B=4\times { 10 }^{ -3 }$
Area vector makes complementary angle with field therefor, angle between magnetic field and normal of coil is $\theta=90^o-60^{ o }={30  }^{ o  }$
$i = 0.2$

$\tau= $$\vec { M } \times \vec { B } $
 $ = niA$$ B\sin\theta$
  $= 500 \times 0.2 \times 400\times 10^{-4} \times 4 \times 10^{-3} \times \dfrac{1}{2} $
$ = 8 \times 10^{-3}Nm$

A rectangular coil of length $10 cm$ and breadth $20 cm$ is placed in uniform magnetic field of induction $20 Wbm^{-2}$ . A current of $2A$ is passed through the coil. If it consists of $100$ turns, the maximum torque experienced 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. $40Nm$

  2. $80Nm$

  3. $4000Nm$

  4. $8000Nm$

Show answer
Correct Option: B
Explanation:

$J= ni AB\sin\theta $
maximum at ,
$ \sin\theta=1$
$\therefore J=niAB=100\times 2\times 20\times 10\times 20\times { 10 }^{ -4 }=80 Nm$

When a coil of area $\overline{\mathrm{A}}$ carrying current $\mathrm{i}$ is suspended in a magnetic field of induction $\overline{\mathrm{B}}$ , then the torque 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. $(\overline{\mathrm{A}} \times \overline{\mathrm{B}})$

  2. $\hat{i}(\overline{\mathrm{A}}-\overline{\mathrm{B}})$

  3. $\hat{i}(\overline{\mathrm{A}} \times \overline{\mathrm{B}})$

  4. $\overline{A}(\hat{i} \times \overline{\mathrm{B}})$

Show answer
Correct Option: C
Explanation:

Torque on magnetic dipole, $\hat i (A\times B)$
where $i$ is current 
$A$ is area vector normal to the area surface
$B$ is magnetic field

A rectangular coil of wire carrying a current is suspended in a uniform magnetic field. The plane of the coil is making an angle of $30^\circ{}$ with the direction of the field and the torque experienced by it is $\tau _{1}$ and when the plane of the coil is making an angle of $60^\circ{}$ with the direction of the field the torque experienced by it is $\tau _{2}$, then the ratio $\tau _{1}$ : $\tau _{2}$ 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. $1:\sqrt{3}$

  2. $\sqrt{3}:1$

  3. $1 : 3$

  4. $3 : 1$

Show answer
Correct Option: B
Explanation:

${ \tau } _{ 1 }=\vec { M } \times \vec { B }  = MB  \sin{ 60 }^{ o  }$
${ \tau } _{ 2 }=\vec { M } \times  \vec { B }  = MB  \sin{ 30 }^{ o }$

If coil is making an angle of $\theta$ with field, then it makes an angle of $(90 -\theta)$ with area vector, i.e. magnetic moment vector.

    $\therefore \quad \dfrac { { \tau } _{ 1 } }{ { \tau } _{ 2 } } =\dfrac { \sqrt { 3 }  }{ 1 } $

A vertical rectangular coil of sides $5 cm \times 2 cm$ has $10$ turns and carries a current of $2A$. The torque(couple) on the coil when it is placed in a uniform horizontal magnetic field of $0.1T$ with its plane perpendicular to the 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. $4\times 10^{-3}N-m$

  2. $0$

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

  4. $10^{-3}N-m$

Show answer
Correct Option: B
Explanation:

If plane is perpendicular to the field, normal vector will be parallel to field 
$\sin \theta=0$
$J = MB \sin\theta = 0$

A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $B$ is in the plane of the coil. If due to current $i$ in the triangle, a torque $\tau $ acts on it. The side $l$ of the triangle 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. $\dfrac { 2 }{ \sqrt { 3 } } \left( \dfrac { \tau }{ Bi } \right) $

  2. $2{ \left( \dfrac { \tau }{ \sqrt { 3 } Bi } \right) }^{ { 1 }/{ 2 } }$

  3. $\dfrac { 2 }{ \sqrt { 3 } } { \left( \dfrac { \tau }{ Bi } \right) }^{ { 1 }/{ 2 } }$

  4. $\dfrac { 1 }{ \sqrt { 3 } } \dfrac { \tau }{ Bi } $

Show answer
Correct Option: B
Explanation:

The current flowing clockwise in an equilateral triangle has a magnetic field in the direction of $\hat { k }$.
$\tau =BiNA\sin { \theta  } $
$\tau =BiNA\sin { { 90 }^{ o } } $
$\tau =Bi\times \dfrac { \sqrt { 3 }  }{ 4 } { I }^{ 2 }\times 1$
[$\because $ Area of equilateral triangle $=\dfrac { \sqrt { 3 }  }{ 4 } { I }^{ 2 }$ and $N=1$]
(assuming)
$\Rightarrow { I }^{ 2 }=\dfrac { 4\tau  }{ \sqrt { 3 } Bi } \Rightarrow I=2{ \left[ \dfrac { \tau  }{ Bi\sqrt { 3 }  }  \right]  }^{ { 1 }/{ 2 } }$