Q. 55

A rectangular loop of sides 20 cm and 10 cm carries a current of 5.0 A. A uniform magnetic field of magnitude 0.20 T exists parallel to the longer side of the loop.

(a) What is the force acting on the loop?

(b) What is the torque acting on the loop?

Answer :


Given-
No. of turns of the coil, n = 50



Magnetic field, B = 0.20 T = 2 × 10−1 T



Magnitude of current, I = 5 A



Length of the loop, l = 20 cm = 20 × 10−2 m,



Breadth of the loop, w = 10 cm = 10 × 10−2 m,



So, area of the loop, A = length ×width = 0.02 m2,


Let ABCD be the rectangular loop.



(a)There is no force acts along the the sides ab and cd, as they are parallel to the magnetic field.




But the force acting on the sides ad and bc are equal in magnitude but opposite


So, they cancel out each other


.
Hence, the net force on the loop is zero.


(b) Torque acting on the coil,




Where


B= applied magnetic field


A= area of rectangular loop


I = current flowing through coil


= angle between the area vector and magnetic field







So, the torque acting on the loop is 0.02 Nm and is parallel to the shorter side


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