Tag: force on current carrying conductor

Questions Related to force on current carrying conductor

A bar magnet has a magnetic moment of $200 A m^2$. The magnet is suspended in a magnetic field of $0.30 N A^{-1} m^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^0$, will be then

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. $30$Nm

  2. $30 \sqrt3$Nm

  3. $60$Nm

  4. $60 \sqrt3$Nm

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Correct Option: A
Explanation:

Torque experienced by a magnet suspended in a uniform magnetic field B is given by
$T = MB sin \theta$
Here, $M = 200 A m^2, B = 0.30 N A^{-1}m^{-1}  and \  \theta = 30^0$
$\therefore T = 200 \times 0.30 \times sin 30^0$
$T = 30 N m$

A magnetic needle kept non parallel to the magnetic field in a non uniform magnetic field experience.

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. A force but not a torque

  2. A torque but not a force

  3. Both a force and a torque

  4. Neither a force nor a torque

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Correct Option: C
Explanation:

Magnetic needle can be considered as a magnetic dipole since it is placed in a non uniform magnetic field it will experience af force as magnetic force on both magnetic force will be different. Also, needle is placed non parallel to the field, so it will experience a torque also.

Therefore, C is correct option. 

A uniform horizontal magnetic field of $7.5\times 10^{-2}$T is set up at angle of $30^o$ with the axis of an solenoid and the magnetic moment associated with it is $1.28$J $T^{-1}$. Then the torque 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. $4.8\times 10^{-2}$N m

  2. $1.6\times 10^{-2}$N m

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

  4. $4.8\times 10^{-4}$N m

Show answer
Correct Option: A
Explanation:

Torque, $\tau =MB\sin\theta$
Here, $M=1.28J$ $T^{-1}$, $B=7.5\times 10^{-2}T, \theta =30^o$
$\therefore \tau =1.28\times 7.5\times 10^{-2}\sin 30^o$
$=1.28\times 7.5\times 10^{-2}\times \dfrac{1}{2}$
$=0.64\times 7.5\times 10^{-2}=4.8\times 10^{-2}$N m.