Consider a square current loop (L = 20 cm) that carries a current I = 6.0 A (see
ID: 2259401 • Letter: C
Question
Consider a square current loop (L = 20 cm) that carries a current I = 6.0 A (see figure below). A constant magnetic field B = 0 .24 T makes an angle theta = 24 degrees with the direction normal to the plane of the loop and perpendicular to the line connecting points A and B.
(a) Find the total magnetic force on the loop.
magnitude:direction:
Consider a square current loop (L = 20 cm) that carries a current I = 6.0 A (see figure below). A constant magnetic field B = 0 .24 T makes an angle theta = 24 degrees with the direction normal to the plane of the loop and perpendicular to the line connecting points A and B. Find the total magnetic force on the loop. Assume the loop is attached to a rotatable rod at points A and B. Find the torque on the loop.
Explanation / Answer
Part A)
The total magnetic force on the loop will be zero. The two sides of the loop that are parallel to the component of the magnetic field will be zero since no force can acts unless the magentic field is perpendicular.
The other two sides will generate a force found by
F = BILsin?, however, since the direction of the current in one of those sides is opposite to the other side, the net force does cancel. However, since the forces are equal but opposite in direction, they do create a torque.
answer: 0
Part B)
The formula for the torque is...
? = BIAsin?
? = (.24)(6)(.2)(.2)(sin 24)
? = .0234 Nm or call it 2.34 X 10-2 Nm if you like