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Velocity of the river with respect to ground is given by `v_0.` Width of the river is d. A swimmer swims (with respect to water) perpendicular to the current with acceleration `a = 2t` (where t is time) starting from rest from the origin O at `t= 0.` The equation of trajectory of the path followed by the swimmer is
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Here, `D = 40.0 m, Delta r= 0.80 m`,
`I = 80 xx 10^(3)A, rho =30 Omega m, R = 4.0 xx 10^(3) Omega`,
As the current spreads uniformly over the hemisphere, the current density at any point on given radius r from the striking point is
`J = I/(2 pi r^(2))`
The magnitude of the electric field at a radial distance r is
`E = rho J = (rho I)/(2 pi r^(2))`
Potential difference between a point at radial distance D and a point `(D + Delta r)` is
`Delta V = - underset(D)overset(Delta + Delta r)(int) E dr = - underset(D)overset(D+Dr)(int) (rho I)/(2 pi r^(2)) dr`
`= (rho I)/(2pi) [1/(D + Delta r) - 1/D] = - (rho I)/(2pi) = (Delta r)/(D (D + Delta r))`
`:.` Current across the swimmere is
`I = (| Delta V|)/R = (rho l)/(2 piR) (Delta r)/(D (D + Delta r))`
`= ( 30 xx (80 xx 10^(3)) xx 0.8)/(2 xx 3.14 xx (4.0 xx 10^(3)) xx 40(40+0.8))`
`=4.68 xx 10^(-2)A`

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