Parallelogram-shape coil pqrs of sides 0.7 m and 0.5 m carries a current of 1.5 a, as shown in fig. It is placed in a magnetic field b 40 t parallel to ps. Find (i) forces on the sides of the coil and (ii) torque on the coil.
Answers
Answer:
Explanation:
(i)
The magnitude of the force on a straight current-carrying conductor of length l situated in a magnetic field B is given by
F=iBl sin Q
Where Q is the angle between the length of the conductor and the field B(vector) measured from the conductor towards B(vector).
Here the magnetic field B(vector) is parallel to the side PS that is, for the sides QP and SR of the coil, Q=60°.
.'. The magnitude of the force F(vector) on the side QP of the coil is
F=1.5A×40N/Am×0.7m×sin60° (=0.866)=36.4N.
According to Fleming's left-hand rule, the force F(vector) on QP will be perpendicular to the plane of the paper, directed upwards.
the force on the side SR will also be 36.4N, but it will be perpendicular to the plane of the paper, directed downwards.
The force on each of the sides PS and RQ will be zero because these sides are parallel to the field B(vector) (Q=0 and 180° respectively).
(ii)
The F(vector) and F(vector) on the sides QP and SR of the coil are equal, parallel and oppositely -directed. Hence, they from a couple whose moment is the torque acting on the coil. The magnitude of the torque is
π (torque)=F×perpendicular distance between F(vector) and F(vector)
=F×PS cos30°=36.4N×0.5m×0.866
=15.76Nm.