A rectangular coil with 250 turns is 6.0 cm long and 4.0 cm wide. When the coil is placed in a magnetic field of 0.25 T, its maximum torque is 0.20 Nm. What is the current in the coil? (Answer: 1.3 A)
Given:
Number of turn = N = 250
Length of the Rectangular Coil = L = 6.0 cm = 0.06 m
Width of the Rectangular Coil = w = 4.0 cm = 0.04 m
Thus Area = A = length x width = 0.06 m x 0.04 m = 0.0024 m²
Magnetic Field = B = 0.25 T
Torque = T = 0.20 Nm
To Find:
The magnitude of the current in the coil = I = ?
Solution:
Since we have the formula for the Torque in a loop
T = N I A B sinθ
For maximum torque angle θ should be 90º and sinθ = sin 90º = 1 So,
T = N I A B
or
I = TNAB
by putting the corresponding values
I = 0.20Nm250x0.0024m²x0.25T
I = 0.20Nm0.15m²T
I = 1.333 A ------------------ Ans.
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