(a) Define the term ‘self inductance’ of a coil. Write its S.I. unit.

(b)  A rectangular loop of sides a and b carrying current I_{2} is kept at a distance ‘a’ from an infinitely long straight wire carrying  current I_{1} as shown in the figure. Obtain an expression for the resultant force acting on the loop.

        

 

 

 

 
 
 
 
 

Answers (1)

Self-inductance (L) is the emf induced in the coil when the rate of change of current in the coil is 1 ampere/second

           \phi =LI

Where \phi= magnetic flux

            I= Current flowing

Where L= coefficient of self- inductance or self-inductance of the coil

when, L=1A

   Then, \phi =L

The S.I unit of self-inductance is Henry (H).

(b) We know the magnetic field due to the infinitely long current-carrying wire is given by :

        B=\frac{\mu _{0}i}{2\pi r}

Where, \mu _{0}=4\pi \times 10^{-7}H/m

              i = current carried by infinite wire

               r= perpendicular distance of wire to that point.

Now, the force acting on a straight wire is given by,

            F=i(l\times B)

For side  AB  

        \vec{F}=i(\vec{l}\times \vec{B})

       F=I_{2}\left ( b\times \frac{\mu _{0}I_{1}}{2\pi a} \right )

       F=\frac{I_{1}I_{2}\mu _{0}b}{2\pi a},

 in -ve x-direction.

Such that, for BC and AD forces are equal but opposite in direction so, they will cancel out each other.

Now, for side CD

    \vec{F}=i(\vec{l}\times \vec{B})

    F=I_{2}\left ( b\frac{\mu _{0}I_{1}}{2\pi (2a)} \right )

    F=\frac{I_{1}I_{2}\mu _{0}b}{4\pi a} in +ve x- direction 

Hence, the net force will be in -ve x-direction ,

    F_{net}=\frac{I_{1}I_{2}\mu _{0}b}{\pi a}\left ( \frac{1}{2}-\frac{1}{4} \right )

    F_{net}=\frac{I_{1}I_{2}\mu _{0}b}{4\pi a}

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