Answer :



𝞍=magnetic flux

I=current flowing

where L=coefficient of self-inductance.

When I= 1A

So, from general understanding, we can say that self-inductance is related to the magnetic flux of a coil when the unit current is flowing through it.

In other words, self-inductance is the emf induced in the coil when the rate of change of current in the coil is 1ampere/second.

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


We know the magnetic field due to an infinitely long current-carrying wire is given by

where H/m

i = current carried by infinite wire

r = perpendicular distance of wire to the point

Force on a straight wire is given by,

Using right-hand rule, we can find the direction of the magnetic field for I1.

According to Fleming's right-hand rule, the thumb, forefinger and middle finger of the right hand are stretched perpendicular to each other as shown in the illustration shown below, and if the thumb represents the direction of the movement of conductor, fore-finger represents direction of the magnetic field, then the middle finger represents direction of the induced current.

In the square magnetic field is acting in the downward direction.

Now, for side AB l is +y direction, B is downward,  is in x-direction and magnitude of the force is,

in –ve x-direction.

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

And for side CD, l is -y-direction, B is downward, is in +x direction and magnitude of the force is,

in +ve x-direction.

So, net force will be in –ve x-direction,

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