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?

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

onyebuchinnaji onyebuchinnaji · Answer 2 · 2022-04-08T17:24:27+02:00

(a) The expression for the magnetic field due to the wire is [tex]B(t) = \frac{2.3\mu_0 t}{2\pi r}[/tex].

(b) The magnitude of the flux through the loop is 3.224 x 10⁻⁸ Tm².

(c) The induced current flowing in the loop is 1.29 x 10⁻⁸ A.

Magnetic field due to the field

The magnetic field at a given distance due to current carrying wire is given as;

[tex]B = \frac{\mu_0 I}{2\pi r}[/tex]

when the current, I is given as a function flow rate of charge, the magnetic field becomes.

I = Qt = 2.3t

[tex]B(t) = \frac{\mu_0 Qt}{2\pi r}= \frac{2.3\mu_0 t}{2\pi r}[/tex]

Magnetic flux through the loop

The magnetic flux through the loop is calculated as follows;

Ф = BA

where;

A is the area of the square loop

A = L² = 0.029 x 0.029 = 0.000841 m²

Relationship between, Q, and 'I'; Q = It ---(1)

I(t) = Qt ---(2)

Substitute (1) into (2)

I = (It)t

1 = t²

t = 1 s

[tex]B(t) = \frac{2.3\mu_0t}{2\pi r} \\\\B(1) = \frac{2.3 \times (4\pi \times 10^{-7}) \times (1)}{2\pi \times (0.012)} \\\\B (1) = 3.833 \times 10^{-5} \ T[/tex]

Ф = BA

Ф = (3.833 x 10⁻⁵) x (0.000841)

Ф = 3.224 x 10⁻⁸ Tm²

Induced emf

The emf induced in the loop is calculated as follows;

[tex]emf = \frac{d\phi}{dt} \\\\emf = \frac{dAB(t)}{dt} \\\\emf = A\frac{dB(t)}{dt} \\\\emf = AB = 3.224 \times 10^{-8} \ V[/tex]

Induced current in the loop

V = IR

[tex]I = \frac{V}{R} \\\\I = \frac{3.224 \times 10^{-8}}{2.5} \\\\I = 1.29 \times 10^{-8} \ A[/tex]

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