Question

A square loop of side length a =2.9 cm is placed a distance b = 1.2 cm from a long wire carrying a current that varies with time at a constant rate, i.e. I(t) = Qt, where Q = 2.3 A/s is a constant. show answer Correct Answer 25% Part (a) Find an expression for the magnetic field due to the wire as a function of time t at a distance r from the wire, in terms of a, b, Q, t, r, and fundamental constants. B(t) = ( μ0 Q t )/( 2 π r ) Part (b) What is the magnitude of the flux through the loop? Select the correct expression. Part (c) If the loop has a resistance of 2.5 Ω, how much induced current flows in the loop?

Answer 1

Answer:

Explanation:

Magnetic field due to a long wire

B =  (  μ₀ / 4π ) x 2 I / r  ( I is current through the wire )

= (μ₀ Q t ) / ( 2 π r )                    [ I = Q t ]

b )  Magnetic field will change through the length of the square shaped loop . So we shall have to take the help of integration to calculate the flux through the loop.

Flux Φ = ∫ B dA

=∫ (  μ₀ / 4π ) x 2 I / r  dA

10⁻⁷ x 2 I ∫ a dr/r

2 x 10⁻⁷ x Qt x a [ ln (2.9+1.2) / 1.2  ]

4.6 x 10⁻⁷ x 2.9 x 10⁻² t  ln 4.1 / 1.2

16.4  x 10⁻⁹ t

Φ( t ) =  16.4 x 10⁻⁹ t

Induced emf  = dΦ( t ) / dt = 16.4 x 10⁻⁹ v

current

16.4 x 10⁻⁹  / 2.5

= 6.56 x 10⁻⁹