Answer:
The current in the coil is 60 Ampere.
Explanation:
Given:
Number of turns in the coil is N = 25
Dimension of the coil = 15cm X 12cm
magnitude of magnetic field = 0.20T
angle in the xy plane is θ = 0 degree
torque τ = 5.4 N-m
To find:
current in the coil is i = ?
Solution:
The torque acting on the coil is given by
=> [tex]\tau = NiAB cos\theta[/tex]
Converting cm to m
12 cm = 0.12 m
15 cm = 0.15 m
The area of the coil is
A = 0.12 X 0.15
A = [tex]0.018 m^2[/tex]
Substituting the values
=>[tex]5.4 = 25\times i \times 0.018 \times 0.20 \times cos\theta[/tex]
=>[tex]5.4 = 25\times i \times 0.018 \times 0.20 \times cos(0)[/tex]
=>[tex]5.4 = 25\times i \times 0.018 \times 0.20 \times 1[/tex]
=>[tex]5.4 = 25\times i \times 0.018 \times 0.20 \times 1[/tex]
=>[tex]5.4 = 0.09\times i[/tex]
=>[tex]i = \frac{5.4}{0.09}[/tex]
=> i = 60 A
