Question

A coil with a circular cross section having an area of has 40turns. When the coil is placed inside a 240cmmagnetic field of 2T, the maximum torque is found to be. (a) Calculate the current in the coil. 6210.xNmi(b)What work is required to rotate the coil by , if the initial orientation of the magnetic dipole moment of 180the loop is at with respect to the field B.

Answer 1

Step-by-step explanation:

cross sectional area A = 40 cm² = 40 x 10⁻⁴ m² , n = 40 , Magnetic field B = 2 T . Current = I .

Maximum torque = 2 x 10⁻⁶ Nm

Value of maximum torque = n B A I .

n B A I = 2 x 10⁻⁶ Nm

40 x 2 x 40 x 10⁻⁴ I = 2 x 10⁻⁶ Nm

.32 I = 2 x 10⁻⁶ Nm

I = 2 x 10⁻⁶ / 32 x 10⁻²

= .0625 x 10⁻⁴

= 6.25 x 10⁻⁶

= 6.25 μ A .

b )

Work done in rotating a coil = n B A I ( cos θ₂ - cosθ₁ )

= n B A I ( cos θ₂ - cosθ₁ )

= 2 x 10⁻⁶ ( cos 180 - cos θ )

= 2 x 10⁻⁶ ( -1 - cos θ )

= - 2 x 10⁻⁶ ( 1 + cos θ ) J .