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Q.(a) State and prove Ampere's circuital law.

(3)
(b) Two long and parallel straight wires A and B, carrying currents of 8.0 A and 5.0 A in the same direction, are seperated by a distance of 4.0 cm. Estimate the force on 10 cm section of wire A. (2)
Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2019Subjective· 5mImportance★★★★★
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(a) ∮B⃗⋅dl⃗=μ0Ienc\oint\vec B\cdot d\vec l=\mu_0 I_{enc}; for a straight wire B=μ0I/2πrB=\mu_0 I/2\pi r. (b) F=μ0I1I22πd l=2×10−5F=\dfrac{\mu_0 I_1 I_2}{2\pi d}\,l = 2\times10^{-5} N, attractive.

(a) Ampere's circuital law — statement. The line integral of the magnetic field B⃗\vec B around any closed loop is equal to μ0\mu_0 times the total current IencI_{enc} passing through (enclosed by) the loop:

∮B⃗⋅dl⃗=μ0Ienc\oint \vec B\cdot d\vec l = \mu_0 I_{enc}

Proof / application to a long straight wire. Consider an infinitely long straight conductor carrying current II. By symmetry the field lines are concentric circles; take an Amperian loop — a circle of radius rr centred on the wire. Everywhere on it B⃗\vec B is tangential and of constant magnitude BB, so

∮B⃗⋅dl⃗=B∮dl=B(2πr)\oint \vec B\cdot d\vec l = B\oint dl = B(2\pi r)

By Ampere's law this equals μ0I\mu_0 I:

B(2πr)=μ0I  ⇒  B=μ0I2πrB(2\pi r) = \mu_0 I \;\Rightarrow\; \boxed{B = \frac{\mu_0 I}{2\pi r}}

(b) Force on a 10 cm section of wire A. Two long parallel wires carrying currents I1I_1 and I2I_2 a distance dd apart exert a force per unit length on each other:

Fl=μ0I1I22πd\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d} …

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