Q. 404.0( 2 Votes )
A circular loop of radius r carrying a current i is held at the center of another circular loop of radius R (>>r) carrying a current I. The plane of the smaller loop makes and angle of 30° with that of the larger loop. If the smaller loop is held fixed in this position by applying a single force at a point on its periphery, what would be the minimum magnitude of the force?
Current carried by circular loop of radius r = i
Current carried by circular loop of radius R(>>r) = I
Angle between the planes of the two loops = 300
Magnetic field at the center of a circular loop of radius r
I = current through the loop,
r = radius of the circle,
𝛍0 = 4π x 10-7 kgm s-2 A-2
Magnetic torque acting on a loop of radius r
I = current carried by the loop,
A = area of loop,
B = magnetic field,
F = force on periphery
Therefore, magnetic field at the center of the loops due to the bigger loop(B) =
Area of the smaller loop(A) = πr2
Therefore, torque on the smaller loop = i (A X B), since the smaller loop carries current i.
A = area of smaller loop, B = magnetic field due to bigger loop.
Now, we know that A X B = |A||B|sinθ,
θ is the angle between the area vector of the smaller loop and the magnetic field of the larger loop.
|A| = magnitude of area of smaller loop,
|B| = magnitude of magnetic field due to larger loop.
The angle between the area vector of the smaller loop and the magnetic field of the larger loop is θ = 300.
Hence, torque acting on the smaller loop = i(|A||B|sin 300) =
We know, |A| = πr2, |B| =
Torque on smaller loop due to magnetic field,
Torque due to external force on its periphery,
F = external force,
r = perpendicular distance between the line of action of force and the point at which the torque is acting.
Here, F and r are perpendicular to each other, since the force is being applied to the periphery.
Here r = radius of smaller loop.
which gives us
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