Question

A wire is formed into a circle having a diameter of 19.0 cm and placed in a uniform magnetic field of 3.40 mT. The wire carries a current of 5.00 A. (a) Find the maximum torque on the wire. µN · m (b) Find the range of potential energies of the wire–field system for different orientations of the circle. minimum µJ maximum µJ

Answer 1

Answer:

Explanation:

Radius = 9.5 x 10⁻² m

area of circle = 3.14 x (9.5 x 10⁻² )²

A = 283.38 x 10⁻⁴ m²

magnetic moment = area x current

M = 283.38 x 10⁻⁴ x 5

= 1416.9 x 10⁻⁴  Am²

Torque = MBsinθ

M is magnetic moment , B is magnetic field .

Max torque = 1416.9 x 10⁻⁴  x 3.4 x 10⁻³ , for θ = 90

= 4817.46 x 10⁻⁷

= 481.7 x 10⁻⁶

= 481.7 μ J

Energy = - MBcosθ

Max  energy  when θ = 180

MB = 4817.46 x 10⁻⁷ J

Min energy = - 4817.46 x 10⁻⁷  for θ = 0

Answer 2

A. Max torque is = 481.7 μ J

B. When The range of potential energies of the wire–field of the circle Min energy is = - 4817.46 x 10⁻⁷ for θ = 0  

Calculation of Potential energies

The Radius is = 9.5 x 10⁻² m

Then,  area of circle is = 3.14 x (9.5 x 10⁻² )²

A is = 283.38 x 10⁻⁴ m²

The magnetic moment is = area x current

M is = 283.38 x 10⁻⁴ x 5

After that, = 1416.9 x 10⁻⁴  Am²

Torque is = MBsinθ

Then, M is magnetic moment, B is magnetic field .

After that, Max torque is = 1416.9 x 10⁻⁴  x 3.4 x 10⁻³ , for θ = 90

= 4817.46 x 10⁻⁷

Then, = 481.7 x 10⁻⁶

= 481.7 μ J

Energy is = - MBcosθ

Max energy  when θ is = 180

After that, MB is = 4817.46 x 10⁻⁷ J

Therefore, Min energy is = - 4817.46 x 10⁻⁷  for θ = 0

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