Answer:
(a) the magnitude of the magnetic moment of the loop is 0.75 Am²
(b) the torque the magnetic field exerts on the loop is 0.028 N.m
Explanation:
Given;
number of turns, N = 15
width of the loop, w = 10 cm = 0.1 m
length of loop, L = 20 cm = 0.2 m
current through the loop, I = 2.5 A
strength of the magnetic field, B = 0.037 T
Area of the loop, A = L x w = 0.2 x 0.1 = 0.02 m²
Part (a) the magnitude of the magnetic moment of the loop
μ = NIA
where;
μ is the magnitude of the magnetic moment of the loop
μ = 15 x 2.5 x 0.02 = 0.75 Am²
Part (b) the torque the magnetic field exerted on the loop
τ = μB
where;
τ is the torque the magnetic field exerts on the loop
τ = μB = 0.75 x 0.037 = 0.028 N.m
Given Information:
Magnetic field = B = 0.037 T
Current = I = 2.5 A
Number of turns = N = 15 turns
Length of rectangular coil = L = 20 cm = 0.20 m
Width of rectangular coil = W = 10 cm = 0.10 m
Required Information:
(a) Magnetic moment = µ = ?
(b) Torque = τ = ?
Answer:
(a) Magnetic moment = 0.75 A.m ²
(b) Torque = 0.0277 N.m
Explanation:
(a) The magnetic moment µ is given by
µ = NIA
Where µ is the magnetic, N is the number of turns, I is the current, A is the area of rectangular loop and is given by
A = W*L
A = 0.10*0.20
A = 0.02 m²
µ = 15*2.5*0.02
µ = 0.75 A.m²
(b) The toque τ exerted on current carrying loop with A area in the presence of a magnetic field B is given by
τ = NIAB
τ = 15*2.5*0.02*0.037
τ = 0.0277 N.m
Alternatively,
τ = µB
τ = 0.75*0.037
τ = 0.0277 N.m
