In the figure below, the cube is 36.0 cm on each edge. Four straight segments of
ID: 1479148 • Letter: I
Question
In the figure below, the cube is 36.0 cm on each edge. Four straight segments of wire—ab, bc, cd, and da—form a closed loop that carries a current I = 4.90 A in the direction shown. A uniform magnetic field of magnitude B = 0.022 T is in the positive y direction. Determine the magnetic force on each segment. (a) ab F = 0 ˆ i +0 ˆ j +0 ˆ k Correct: Your answer is correct. mN (b) bc F = 38.8 ˆ i +0 ˆ j +0 ˆ k Correct: Your answer is correct. mN (c) cd F = 0 ˆ i +0 ˆ j 27.4 ˆ k Incorrect: Your answer is incorrect. mN (d) da F = 54.9 ˆ i +0 ˆ j 54.9 ˆ k Incorrect: Your answer is incorrect. mN
Explanation / Answer
side length of the cube, i=36cm,
curent i=4.9 A
magnetic field B=0.022T
magnetic force on a segment of wire, F=i*L*B*sin(theta)
a)
on the segment ab:
F=i*L*B*sin(theta)
F=i*l*B*sin(theta)
F=4.9*0.36*0.022*sin(180)
F=0
or,
F=0i+0j+0k
b)
on the segment bc:
F=i*L*B*sin(theta)
F=i*l*B*sin(theta)
F=4.9*0.36*0.022*sin(90)
F=38.8 mN
or
F=((-38.8)i+0j+0k) mN
c)
on the segment cd:
F=i*L*B*sin(theta)
F=i*l*sqrt(2)*B*sin(45)
F=4.9*0.36*sqrt(2)*0.022*sin(45)
F=38.8 mN
or
F=(0)i+0j+((-38.8)k) mN
d)
on the segment da:
F=i*L*B*sin(theta)
F=i*l*sqrt(2)*B*sin(90)
F=4.9*0.36*sqrt(2)*0.022*sin(90)
F=54.9 mN
or
F=(54.9)i+0j+((54.9)k) mN