deduce the expression for the torque experienced by a rectangular loop carrying a Steady Current I and place in uniform magnetic field b
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The length of loop = PQ = RS= l and breadth = QR = SP = b. Let at any instant the normal to the plane of loop make an angle q with the direction of magnetic field vector B and I be the current in the loop. We know that a force acts on a current carrying wire placed in a magnetic field. Therefore, each side of the loop will experience a force. The net force and torque acting on the loop will be determined by the forces acting on all sides of the loop. Suppose that the forces on sides PQ, QR, RS and SP are vector F1, vector F2, vector F3 and vector F4 respectively. The sides QR and SP make angle (90° - θ) with the direction of magnetic field. Therefore each of the forces vector F2 and vector F4 acting on these sides has same magnitude F' = Blbsin(90° - θ) = Blbcosθ. According to Fleming’s left hand rule the forces F2 and vector F4 are equal and opposite but their line of action is same. Therefore these forces cancel each other i.e. the resultant of vector F2 and vector F4 is zero. The sides PQ and RS of current loop are perpendicular to the magnetic field, therefore the magnitude of each of forces vector F1 and vector F3 is
F = IlBsin90° = IlB.
According to Fleming’s left hand rule the forces vector F1 and vector F3acting on sides PQ and RS are equal and opposite, but their lines of action are different; therefore the resultant force of vector F1 and vector F3 is zero, but they form a couple called the deflecting couple. When the normal to plane of loop makes an angle θ with the direction of magnetic field B, the perpendicular distance between vector F1 and vector F3 is bsinθ.
The length of loop = PQ = RS= l and breadth = QR = SP = b. Let at any instant the normal to the plane of loop make an angle q with the direction of magnetic field vector B and I be the current in the loop. We know that a force acts on a current carrying wire placed in a magnetic field. Therefore, each side of the loop will experience a force. The net force and torque acting on the loop will be determined by the forces acting on all sides of the loop. Suppose that the forces on sides PQ, QR, RS and SP are vector F1, vector F2, vector F3 and vector F4 respectively. The sides QR and SP make angle (90° - θ) with the direction of magnetic field. Therefore each of the forces vector F2 and vector F4 acting on these sides has same magnitude F' = Blbsin(90° - θ) = Blbcosθ. According to Fleming’s left hand rule the forces F2 and vector F4 are equal and opposite but their line of action is same. Therefore these forces cancel each other i.e. the resultant of vector F2 and vector F4 is zero. The sides PQ and RS of current loop are perpendicular to the magnetic field, therefore the magnitude of each of forces vector F1 and vector F3 is
F = IlBsin90° = IlB.
According to Fleming’s left hand rule the forces vector F1 and vector F3acting on sides PQ and RS are equal and opposite, but their lines of action are different; therefore the resultant force of vector F1 and vector F3 is zero, but they form a couple called the deflecting couple. When the normal to plane of loop makes an angle θ with the direction of magnetic field B, the perpendicular distance between vector F1 and vector F3 is bsinθ.
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A current of electricity is a steady flow of electrons. When electrons move from one place to another, round a circuit, they carry electrical energy from place to place like marching ants carrying leaves. Instead of carrying leaves, electrons carry a tiny amount of electric charge.
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