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 ï½˜ 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 = `\frac {T}{NAB}`

by putting the corresponding values

I = `\frac {0.20 Nm}{250 ｘ 0.0024 m²ï½˜ 0.25 T}`

I = `\frac {0.20 Nm}{0.15 m² T}`

I = 1.333 A ------------------ Ans.



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