Q. 655.0( 3 Votes )

The current in an ideal, long solenoid is varied at a uniform rate of 0.01 A s–1. The solenoid has 2000 turns/m and its radius is 6.0 cm.

(a) Consider a circle of radius 1.0 cm inside the solenoid with its axis coinciding with the axis of the solenoid. Write the change in the magnetic flux through this circle in 2.0 seconds.

(b) Find the electric field induced at a point on the circumference of the circle.

(c) Find the electric field induced at a point outside the solenoid at a distance 8.0 cm from its axis.

Answer :

Given:


Rate of variation of current = 0.01 A s–1.


No of turns/m (n) = 2000


Radius(r) = 6 cm = 0.06 m


Formula used:


(a) Radius of circle(r’) = 1 cm = 0.01 m


Time(t) = 2 s


For two seconds, change of current = (2 x 0.01 A.) = 0.02 A


Magnetic flux , where B = magnetic field, A = area


Area of circle


Magnetic field of a solenoid , where μ0 = magnetic permeability of vacuum, n = number of turns per unit length, Δi = change in current


Hence, flux =


Wb


Hence, in 1 second = Wb (Ans)


(b) = , where E = electric field, dr = line element, E’ = emf, = flux, t = time


Hence, in this case, this becomes


, where r = radius of circle


Vm-1(Ans)


(c) For the point located outside,


Wbs-1


= flux, t = time, μ0 = magnetic permeability of vacuum, n = number of turns per unit length, di/dt = rate of change in current


, where E = electric field, r = radius of circle(since )


Hence,


Vm-1 (Ans)


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