(a) Derive the expression for the force acting between two long parallel current carrying conductors. Hence, define 1 A current.

(b) A bar magnet of dipole moment 3 Am2 rests with its centre on a frictionless pivot. A force F is applied at right angles to the axis of the magnet, 10 cm from the pivot. It is observed that an external magnetic field of 0.25 T is required to hold the magnet in equilibrium at an angle of 30^{\circ} with the field. Calculate the value of F. How will the equilibrium be effected if F is withdrawn ?

 

 
 
 
 
 

Answers (1)
S safeer

a)

B_{1}=\frac{\mu_{o}I_{1}}{2\pi d}

B_{2}=\frac{\mu_{o}I_{2}}{2\pi d}

F_{12}=I_{2}B_{1}L

F_{21}=I_{1}B_{2}L

F = F_{12}=F_{21}=\frac{\mu_{o}II_{2}L}{2\pi d}

\frac{f}{L}=\frac{\mu_{o}I_{1}I_{2}}{2\pi d}

Ampere is that value of steady-state current, which when maintained in each of the very two long, straight, parallel conductors of negligible cross-section and placed 1m apart in a vacuum produces 2\times 10^{-7}\; N/m force on each conductor.

b) At equilibrium
Restoring Torque = Deflecting Torque

F\ \times\ r=mBsin\theta

\\F\times \10\times10^{-2}=3\times0.25\times sin30\\F=3.75N

The magnet oscillates for some time but finally aligns along the original direction of the external magnetic field.

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