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Consider a circular current-carrying loop of radius R in the x-y plane with centre at origin. Consider the line integral (L)=|int_(-L)^(L)vecB*vec(dl)| taken along z-axis. (a) Show that (L) monotonically increases with L. (b) Use an appropriate Amperian loop to show that (oo)=mu_(0)I, where I is the current in the wire. ( c) Verify directly the above result. (d) Suppose we replace the circular coil by a square coil of sides R carrying the same current I. What can you say about (L) and (oo)? |
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Answer» Solution :(a) B(z) is same at every point on z-axis. So, is monotonically function for L. `vecBandvec(dl)` are in same direction. So, `VECB*vec(dl)=Bdlcos0=Bdl` (b) (L) + by circumference at large distance `C=mu_(0)I` Now, `Ltooo` At large distance `TO0" "(becauseBprop1/r^(3))` `(oo)-mu_(0)(I)` ( c)  Current CARRYING ring of radius R is in xy-plane. Centre of it is at origin point, magnetic field at any point from centre of loop is, `B_(z)=(mu_(0)IR^(2))/(2(z^(2)+R^(2))^(3/2))` `int_(-oo)^(oo)B_(z)dx=int_(-oo)^(oo)(mu_(0)IR^(2))/(2(z^(2)+R^(2))^(3/2))dz` Taking `z=Rtantheta` `dz=Rsec^(2)thetad theta` `int_(-oo)^(oo)B_(z)dz=(mu_(0)I)/2int_((-pi)/2)^((+pi)/2)(costheta)d theta=mu_(0)I` (d) `B(z)_("square")ltB(z)_("circular loop")` `thereforeJ(L)_("square")ltJ(L)_("circular loop")` but as per DISCUSSION of option (b), `J(oo)_("square")=J(oo)_("circular loop")` |
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