Tag: torque on current carrying loop

Questions Related to torque on current carrying loop

A bar magnet of magnetic moment 1.5 J/T is along the direction of the uniform magnetic field of 0.22T. The work done in turning the magnet opposite to the field direction and the torque required to keep in that position are 

  1. $0.33J$ and $0.33 N-m$

  2. $0.66J$ and $0.66 N-m$

  3. $0.33J$ and $0 N-m$

  4. $0.66J$ and $0 N-m$


Correct Option: D

On applying a uniform magnetic field on a current-carrying coil the coil rotates in such a way that its plane

  1. becomes perpendicular to magnetic field

  2. becomes parallel to magnetic field

  3. makes an angle of $45^o$ with the magnetic field

  4. makes any angle with the magnetic field


Correct Option: A
Explanation:

On applying a uniform magnetic field on a current-carrying coil, the lines of force are at right angle to the plane of coil. Hence, the coil rotates in such a way that its plane becomes perpendicular to magnetic field.

A very long magnet of pole strength 16 A-m is placed vertically with its one pole on the table. At what distance from the pole, there will be a neutral point on the table. $(B _H =4 \times 10^{-5} \ Wbm^{-2})$

  1. 0.4 m

  2. 0.2 m

  3. 0.5 m

  4. 0.8 m


Correct Option: A

The torque $(\vec t)$ experienced by a current - loop of magnetic moment $(\vec M)$ placed in magnetic field $\vec B$ is -

  1. $\vec t = \vec M \times \vec B$

  2. $\vec t = \vec B \times \vec M$

  3. $\vec t = \frac{\vec M}{\vec B}$

  4. $\vec t = \vec M.\vec B$


Correct Option: A

A coil of area 0.01 m$^2$ is lying in a perpendicular magnetic field of 0.1 Tesla. If a current of 10 A is passed in it then the maximum torque acting on the coil will be

  1. 0.01 N/m

  2. 0.001 N/m

  3. 1.1 N/m

  4. 0.8 N/m


Correct Option: A
Explanation:

Magnetic moment = $ I \vec A = 0.01 \times 10   A/m^2$ perpendicular to the field.
Maximum torque on the magnetic moment is when angle between magnetic moment and the field is $90^{\circ}= (I \vec A ) \times \vec B = I A B = 10 \times 0.01 \times 0.1 = 0.01 Nm $

A flat coil carrying a current has a magnetic moment $\vec{\mu}$. It is placed in a magnetic field $\vec B$. The torque on the coil is $\vec{\tau}$

  1. $\vec{\tau} = \vec{\mu} \times \vec B$

  2. $\vec{\tau} = \vec{B} \times \vec{\mu} $

  3. $|\vec{\tau}| = \vec{\mu} \cdot \times \vec B$

  4. $\vec{\tau}$ is perpendicular to both $\vec{\mu}$ and $\vec{B}$.


Correct Option: A,D
Explanation:

The magnetic moment is defined as a vector relating the aligning torque on the object from an externally applied magnetic field to the field vector itself. The relationship is given by:

$ \tau = \vec{\mu} \times \vec{B} $

where  $\tau$ is the torque acting on the dipole and $B$ is the external magnetic field, and $\mu$  is the magnetic moment. Direction of torque is given by the right hand rule.

A current-carrying loop suspended freely in a uniform magnetic field  will experience 

  1. torque only

  2. force only

  3. neither torque nor force

  4. both


Correct Option: A
Explanation:

A current carrying loop behaves as a magnetic dipole. and we know that a dipole placed in uniform magnetic field only experiences torque.

Asteady current 'I' flows in a small square loop of wire of side $L$ in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let $\overline{\mu} _1$ and $\overline{\mu} _2$ respectively denote the magetic moments of the current loop before and after folding. Then:

  1. $\overline{\mu} _2 = 0$

  2. $\overline{\mu} _1$ and $\overline{\mu} _2$ are in the same direction

  3. $\dfrac{|\overline{\mu} _1|}{|\overline{\mu} _2|} = \sqrt{2}$

  4. $\dfrac{|\overline{\mu} _1|}{|\overline{\mu} _2|} = \dfrac{1}{\sqrt{2}}$


Correct Option: C

A current carring wire is bent to from a circuital coil. If this coil is placed in any other magnetic filed the maximum torque on the coil, the number of turns will be 

  1. 1

  2. 2

  3. 4

  4. 8


Correct Option: C

The plane of a rectangular loop of wire with sides $0.05 m$ and $0.08 m$ is parallel to a uniform magnetic field of induction $1.5\times 10^{-2}T$ . A current of $10.0 A$ flows through the loop. If the side of length $0.08 m$ is normal and the side of length $0.05 m$ is parallel to the lines of field, then the torque acting on it is

  1. $6000N-m$ 

  2. $zero$ 

  3. $1.2\times 10^{-2}N-m$

  4. $6\times 10^{-4}N-m$ 


Correct Option: D
Explanation:

Torque on the loop $\vec{\tau} = \vec{\mu} \times \vec{B}$ where $\vec{\mu}=i\vec{A}$ is the magnetic moment of the loop and $\vec{B}$  is the magnetic field.
$ \therefore \tau = i\vec{A} \times \vec{B}$
Since $\vec{B}$ is in the plane of the loop, $\vec{A} \perp \vec{B}$
$ \therefore \vec{\tau} = 10 \times ( 0.05 \times 0.08) \times( 1.5 \times 10^{-2})= 6 \times 10^{-4}N-m$