Tag: force on current carrying conductor

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Questions Related to force on current carrying conductor

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 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

  1. $30$Nm

  2. $30 \sqrt3$Nm

  3. $60$Nm

  4. $60 \sqrt3$Nm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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. 

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 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?

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.