A rectangular wire-loop of width a is suspended from the insulated pan of a spring balance as shown in figure. A current i exists in the anticlockwise direction in the loop. A magnetic field B exists in the lower region. Find the change in the tension of the spring if the current in the loop is reversed.
Given-
width of wire loop = a
Electric current through the loop = i
Direction of the current is anti-clockwise.
Strength of the magnetic field in the lower region = B
From fig, we can say that direction of the magnetic field is into the plane of the loop.
Here, angle between the length of the loop and magnetic field, θ = 900
Magnetic force is given by
where,
B= magnetic field
I = current
l = length of the wire
and θ = the angle between B and l
The magnetic force will act only on side AD and BC.
As side AD is outside the magnetic field, so F = 0
Magnetic force on side BC is
Direction of force can be found using Fleming’s left-hand rule.
whenever a current carrying conductor is placed inside a magnetic field, a force acts on the conductor, in a direction perpendicular to both the directions of the current and the magnetic field
Thus, the direction of the magnetic force will be upward.
Similarly altering the direction of current to clockwise, the force
along BC-
Thus, the change in force is equal to the change in tension
