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.

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 .

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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