(a) Define mutual inductance and write its S.I. unit.
(b) A square loop of side ‘a’ carrying a current I_{2} is kept at distance x from an infinitely long straight wire carrying a current I_{1} as shown in the figure. Obtain the expression for the resultant force acting on the loop.

 
 
 
 
 

Answers (1)

(a) The property of two coils that opposes the change in current in one coil is defined as mutual Inductance. Let, the two coils :- coil 1 with number of turns N_{1} and coil 2 with the number of turns N_{2}.

The magnetic flux across them is \phi _{1} and \phi _{2} respectively.

Let I_{1} be the current through coil 1 and I_{2} be the current through coil 2.

Then, the mutual inductance (m) is given by:

M=\frac{N_{1}\phi _{1}}{I_{2}}=\frac{N_{2}\phi _{2}}{I_{2}}

The SI unit is Henry(H).

 

(b) Consider that two straight infinite parallel conducting wires with current I_{1} and I_{2} are separated by a distance x', then force per unit length of each wire is given by ;

F=\frac{\mu _{o}I_{1}I_{2}}{2\pi x} where, \mu _{o} is the magnetic permeability of free space.

such that, when the wires carry currents in the opposite direction, they repel and when the wires carry currents in the same direction, they attract.

Such that, F_{1}  and F_{2} must be equal. So, they can cancel out.

Now,

\frac{F_{3}}{a}=-\frac{\mu _{o}I_{1}I_{2}}{2\pi (x+a)}

\frac{F_{4}}{a}=\frac{\mu _{o}I_{1}I_{2}}{2\pi x}

Hence, the total force acting on the loop will be -

F = F_{3}+F_{4}

F=\frac{-\mu _{o}I_{1}I_{2}}{2\pi (x+a)}a+\frac{\mu _{o}I_{1}I_{2}}{2\pi x}a

F=\frac{\mu _{o}I_{1}I_{2}a}{2\pi}\left ( \frac{1}{x}-\frac{1}{x+a} \right )

F=\frac{\mu _{o}I_{1}I_{2}a^{2}}{2\pi x (x+a)}

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