A square current carrying loop is suspended in a uniform magnetic fiel...
When a current carrying loop is placed in a magnetic field, the coil experiences a torque given by T = NBiA sin θ. Torque is maximum when θ =90, i.e. the plane of the coil is parallel to the field Tmax NBiA
Force stack bold F subscript bold 1 with bold rightwards arrow on top and bold F with bold rightwards arrow on top subscript 2 acting on the coil are equal in magnitude and opposite in direction. As the forces stack bold F subscript bold 1 with bold rightwards arrow on top and bold F with bold rightwards arrow on top subscript 2 have the same line action their reultant effect on the coil is zero.
The two forces stack bold F subscript bold 3 with bold rightwards arrow on top bold space bold and bold space stack bold F subscript bold 4 with bold rightwards arrow on top
are equal in magnitude and opposite in direction. As the two forces have different lines of action, they constitute a torque .Thus, if the force on one arc of the loop is bold F with bold rightwards arrow on top, the net force on the remaining three arms of the loop is -F.
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A square current carrying loop is suspended in a uniform magnetic fiel...
Introduction:
When a square current-carrying loop is suspended in a uniform magnetic field, each arm of the loop experiences a force due to the interaction between the magnetic field and the current. The force on one arm of the loop can be denoted as F.
Explanation:
Force on one arm of the loop:
Let's consider one arm of the square loop. According to the right-hand rule, the force on a current-carrying wire in a magnetic field is given by the equation F = I * B * L * sin(θ), where I is the current, B is the magnetic field, L is the length of the wire, and θ is the angle between the wire and the magnetic field.
In the case of a square loop, the length of each arm is the same, so we can consider L as a constant value. The angle θ is 90 degrees because the magnetic field acts in the plane of the loop. Therefore, the force on one arm of the loop can be represented as F = I * B * L.
Net force on the remaining three arms:
Since the magnetic field acts in the plane of the loop, the magnetic field lines are perpendicular to the loop. As a result, the forces on the remaining three arms of the loop are also perpendicular to those arms.
Vector representation:
To find the net force on the remaining three arms, we need to consider the vector representation of the forces. Since the forces are perpendicular to the arms, they can be represented as vectors along the three arms, pointing in the same direction.
Equal magnitude and direction:
The forces on the remaining three arms have the same magnitude as the force on the first arm (F), and they all act in the same direction due to the perpendicularity of the forces and the arms. Therefore, the net force on the remaining three arms is three times the force on one arm, which can be expressed as 3F.
Conclusion:
In conclusion, when a square current-carrying loop is suspended in a uniform magnetic field acting in the plane of the loop, the force on one arm of the loop is represented as F. The net force on the remaining three arms of the loop is three times the force on one arm, which can be expressed as 3F.
A square current carrying loop is suspended in a uniform magnetic fiel...
-F becz. Net force is zero
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