Question

A wire is formed into a circle having a diameter of 10.0cm and is placed in a uniform magnetic field of 3.00mT . The wire carries a current of 5.00 A. Find (b) the range of potential energies of the wire-field system for different orientations of the circle.

Answer 1

The range of potential energies of the wire-field system for different orientations of the circle are -

θ                  U

0°             375 π x $10^{-7}$

90°              0

180°        - 375 π x $10^{-7}$

We have current carrying wire in a form of a circle placed in a uniform magnetic field.

We have to the range of potential energies of the wire-field system for different orientations of the circle.

What is the formula to calculate the Magnetic Potential Energy?

The formula to calculate the magnetic potential energy is -

U = M.B = MB cos $\theta$

where -

M is the Dipole Moment.

B is the Magnetic Field Intensity.

According to the question, we have -

U = M.B = MB cos $\theta$

We can write M = IA (I is current and A is cross sectional Area)

U = IAB cos $\theta$

U = Iπ$r^{2}$B cos $\theta$

For $\theta$ = 0° →

U(Max) = MB cos(0) = MB =  Iπ$r^{2}$B = 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

375 π x $10^{-7}$.

For $\theta$ = 90° →

U = MB cos (90) = 0

For $\theta$ = 180° →

U(Min) = MB cos(0) = - MB =  - Iπ$r^{2}$B = - 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

- 375 π x $10^{-7}$.

Hence, the range of potential energies of the wire-field system for different orientations of the circle are -

θ                  U

0°             375 π x $10^{-7}$

90°              0

180°        - 375 π x $10^{-7}$

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