1. A horizontal wire is hung from the ceiling of a room by two massless strings.
ID: 3278141 • Letter: 1
Question
1. A horizontal wire is hung from the ceiling of a room by two massless strings. The wire has a length of
0.23 m
and a mass of
0.077 kg.
A uniform magnetic field of magnitude
0.069 T
is directed from the ceiling to the floor. When a current of
I = 42 A
exists in the wire, the wire swings upward and, at equilibrium, makes an angle with respect to the vertical, as the drawing shows.
(a) Find the angle .
°
(b) Find the tension in each of the two strings.
N
2. Two circular coils of current-carrying wire have the same magnetic moment. The first coil has a radius of 0.084 m, has 135 turns, and carries a current of 3.7 A. The second coil has 171 turns and carries a current of 9.1 A. What is the radius of the second coil?
m
Explanation / Answer
1. given length of the wire, l = 0.23 m
mass of wire , m = 0.077 kg
Magnetic field magnitude, B = 0.069 T
current in wire, i = 42 A
let the angle sat which equilibrium is obtained be theta
let thensio in the strings be T
magnetic force = F = Bil ( towards right)
from force balance we get
Tcos(phi) = mg
Tsin(phi) = Bil
so, tan(phi) = Bil/mg = 0.069*42*0.23/0.077*9.81
phi = 41.425 deg
T = mg/cos(phi) = 0.077g/cos(41.425) = 1.007 N
so tension in each wire, T/2 = 0.503 N
2. magnetic moment of a current carrying coil = ni*A ( i is the current , A is the cross sectional area of the coil)
so, r1 = 0.084 m
n1 = 135
i1 = 3.7
n2 = 171
i2 = 9.1
r2 = ?
so, 135*3.7*0.084^2 = 171*9.1*r2^2
r2 = 0.047 m
3. given length of wire, l = 0.07 m
current in wire, i = 4.3 A
Magnetic field, B = 3.3 T
Maximum torque on a loop of current carryiong wire = BINA
A = pi*r^2
but 2*pi*r = 0.07
r = 0.01114 m
A = 3.899*10^-4 m^2
so T = 3.3*4.3*A = 0.005533 Nm
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