Q. 39

Acircular loop of radius R carries a current I. Another circular loop of radius r (<<R) carries a current i and is placed at the center of the larger loop. The planes of the two circles are at right angle to each other. Find the torque acting on the smaller loop.

Answer :

Given:


Current carried by circular loop of radius R = I


Current carried by circular loop of radius r(<<R) = i


Diagram:



Formula used:


Magnetic field at the center of a circular loop = 𝛍0I/2r,


where 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 (A X B), where I = current carried by the loop, A = area of loop, B = magnetic field


Therefore, magnetic field acting at the center of the loops due to the bigger loop of radius R(B) = μ0I/2R.


Area of the smaller loop(A) = πr2


Hence, torque acting on the smaller loop = I (A X B), since current carried by smaller loop is i. Here A = area of smaller loop = πr2 and B = magnetic field due to bigger loop.


Now, we know that A X B = |A||B|sinθ, where θ is the angle between the area vector of the smaller loop and the magnetic field of the larger loop. |A| = modulus value of area of smaller loop, |B| = modulus value of magnetic field due to larger loop.


Since the planes of the two loops are perpendicular to each other, θ = 900.


Hence, torque acting on the smaller loop = i (|A||B|sin 900) = i(|A||B|)


We know, |A| = πr2, |B| =


Hence, torque on smaller loop = = (Answer)


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