Question

A current of 10 A is flowing in a long straight wire situated near a rectangular coil. The two sides, of the coil, of length 0.2 m are parallel to the wire. One of them is at a distance of 0.05 m and the other is at a distance of 0.10 m from the wire. The wire is in the plane of the coil. Calculate the magnetic flux through the rectangular coil. If the current decays uniformly to zero in 0.02 s, find the emf induced in the coil and indicate the direction in which the induced current flows.

Solution

Given, a rectangular coil and a straight wire carrying current of 10 A.
Sides of the coil = 0.2 m
Distance of one side of coil from the wire = 0.05 m
Distance of second side of coil from the wire = 0.10 m
Time taken to decay the current to 0A = 0.02 s

Consider a strip of width 'dr' at a distance 'r' from the straight wire.
Magnetic field at the location of the strip due to the wire,

B = \frac{μ_{0} I}{2 πr}

Area of strip, dA = l.dr

Magnetic flux linked with the strip,

{dϕ}_{B} = BdA = \frac{μ_{0} I}{2 πr} ldr

Total magnetic flux linked with the coil,

{dϕ}_{B} = \frac{μ_{0} I l}{2 π} \frac{dr}{r}

\int d φ_{B} = \frac{μ_{0} I l}{2 π} \int_{r_{1}}^{r_{2}} \frac{d l}{r}

φ_{B} = \frac{μ_{0} I l}{2 π} [\log_{e} r_{2} - \log_{e} r_{1}] φ_{B} = \frac{μ_{0} I . l}{2 π} {[\log r]}_{r_{1}}^{r_{2}} φ_{B} = \frac{μ_{0} I l}{2 π} \log_{e} \frac{r_{2}}{r_{1}}

Substituting values, we get

ϕ_{B} = \frac{4 π \times 10^{- 7} \times 10 \times 0.2}{2 π} \log [\frac{0.10}{0.05}]

= 4 \times 10^{- 7} \log_{e} 2 = 4 \times 0.693 \times 10^{- 7} Wb = 2.77 \times 10^{- 7} Wb

Induced e.m.f,

|E| = - \frac{{dϕ}_{B}}{dt} = \frac{2.77 \times 10^{- 7}}{0.02} V = 1.39 \times 10^{- 5} V

Magnetic field, due to wire, at the location of the coil is perpendicular to the plane of the coil and directed inwards. When current is reduced to zero, this magnetic field decreases. To oppose this decrease, induced current shall flow clockwise, so that its magnetic field is perpendicular to the plane of the coil and downward.

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