Torque experienced by a current carrying loop in a uniform magnetic field

Magnetic Field

A magnetic field is a vector field that describes the magnetic effect on moving electric charges, electric currents and magnetic substances. A mobile charge in a magnetic field experiences a force perpendicular to the velocity of the mobile charge and to the magnetic field.

 A permanent magnet’s magnetic field pulls on ferromagnetic substances along with iron, and attracts or repels different magnets. In addition, a magnetic field that varies with area will exert a force on a variety of non-magnetic substances through affecting the movement in their outer atomic electrons. Magnetic fields surround magnetised substances, and are created by electric powered currents along with the ones utilised in electromagnets, and by electric powered fields varying in time.

Fleming’s Right Hand Thumb Rule

It states that if we stretch our index finger, middle finger and the thumb in any such manner that they’re together perpendicular to every other in which the index finger shows the route of the magnetic field, middle finger, the route of an precipitated current, at the same time the thumb represents the route of motion.

As we align our fingers in this way, we will see that the current and the magnetic field are in contrary directions. So, they are parallel, i.e sin 00 = 0. Therefore, no force is being experienced by the wire CD. Similarly, if we study the wire AB, the path of current is contrary to that withinside the wire CD, but withinside the identical path to that of the magnetic field.

Similarly,sin 00 = 0 , no force is appearing on wire AB. For wire CA, The current is flowing in an upward path, because of this the electrons are flowing withinside the contrary path. Applying Fleming’s right-hand rule, the magnetic field is in a path perpendicular to that of current. The force is appearing inwards. While for the wire BD, the path of the current is downward and the path of force is outward. A torque is exerted at the loop about an axis, making the loop rotate.

Torque in a current carrying loop Equation

Let us consider a rectangular loop carrying a consistent current I and positioned in a uniform magnetic field experiencing a torque. It does not enjoy a net force. This behaviour has similarities to that of an electrical dipole in a uniform electrostatic field.

We first take the easy case while the rectangular loop is positioned such that the uniform magnetic field B is withinside the plane of the loop. The magnetic field exerts no force on the 2 hands AD and BC of the loop. It is perpendicular to the arm AB of the loop and exerts a force F1 on it that is directed into the plane of the loop. Its value is,

F1=IlB

Similarly, it exerts a force F2 at the arm CD and F2is directed out of the plane of the paper. 

F2=IlB=F1

Thus, the net force at the loop is zero. There is a torque at the loop because of the pair of forces F1and F2. The magnitude of this torque is,

=F1a/2+F2a/2

=IlB(a/2)+IlB(a/2)

=I(ab)B

=IAB……..(1)

Here, A=ab is the area of the rectangle.

Now we take the case while the plane of the loop, isn’t alongside the magnetic field, however makes an angle with it. We take the angle among the sector and the normal to the coil to be angle (The first case corresponds to =/2 ).

The forces at the hands BC and DA are equal, opposite, and act alongside the axis of the coil, which connects the centres of mass of BC and DA. Being collinear alongside the axis they cancel each other, as a result of which there is no net force or torque. The forces on hands AB and CD are F1 and F2. They too are equal and opposite, with magnitude,

F2=F1=IlB

But they are not collinear! This outcomes in a couple as before. The torque is, however, much less than first case whilst plane of loop was alongside the magnetic field. This is due to the fact the perpendicular distance among the forces of the couple has decreased.

The value of torque on the loop can be given by,

=F1a/2 sin+F2a/2 sin

=IAB sin

=IAB sin………(2)

As 0 , the perpendicular distance between the forces of the couple additionally becomes zero. This makes the forces collinear and the net pressure and torque zero. The torques in Eqs. (1) and (2) may be expressed as vector made from the magnetic moment of the coil and the magnetic field. We express the magnetic moment of the current loop as,

m=IA……..(3)

Here, the direction of the area vector A can be  given by fleming’s right-hand thumb rule and it is directed into the plane of the sheet. Then since the angle among m and B is , Eqs. (1) and (2) can be written as,

=mB……..(4)

Conclusion

Torque experienced by a current carrying loop in uniform magnetic field is  given by
=mB
Where m is magnetic moment of loop B is a magnetic field.
Force is zero on a current carrying loop in a uniform magnetic field.
Note : If we want the net force to be zero, the magnetic field should be uniform
When the magnetic field is not uniform, net force may not be zero.

The torque experienced by a current carrying loop of any shape in uniform magnetic field is given by,

=NIAB sin

= Torque

N = number of turns

I =Current

A = Area of the loop

B = Magnetic field

= Angle among the normal to the loop and the uniform magnetic field