lona asked in Science & MathematicsPhysics · 5 months ago

 Flux due to wire ...What is the magnetic flux through the loop at time t= 2.00 seconds? (Define positive flux into the page.) please helpp?

A rectangular loop with sides of length a= 1.30 cm and b= 3.70 cm is placed near a wire that carries a current that varies as a function of time: it=3.67 + 1.12t2 where the current is in Amperes and the time is in seconds. The distance from the straight wire to the closest side of the loop is d= 0.420 cm.

a)What is the magnetic flux through the loop at time t= 2.00 seconds? (Define positive flux into the page.)b)What is the induced e.m.f. in the loop at time t= 2.00 seconds? (Note that positive emf is clockwise.)

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  • ?
    Lv 7
    5 months ago

    Thank you for Best Answer to the other version of this question!  I can't edit it now, so here's my worked solution.

    __________________

    A.  You can’t just multiply the area by B to get the flux because B is not constant across the area. And you can’t use a simple average because the field does not vary linearly with distance from wire. B is stronger nearer the wire according to B = μ₀i/(2πr).  Integration is needed.

    We break the area into strips parallel to the wire.  Each strip is length 0.037m and width dr, so area of each strip is  dA=0.037dr.  Flux through the strip is BdA:

    dΦ = BdA

    = (μ₀i/(2πr)) 0.037dr

    = (4π*10^-7i/(2πr)) * 0.037dr

    = 7.4*10^-9 i dr/r

    To get total flux through loop, we integrate dΦ from r = 0.0042m to r = 0.00420+0.0130 = 0.0172m

    Φ = ∫dΦ

    = 7.4*10^-9 i ∫r/dr (for r=0.0042m to 0.0172m)

    = 7.4*10^-9 i [ln(r)]

    = 7.4*10^-9 i [ln(0.0172) – ln(0.0042)]

    = 7.4*10^-9 i [1.4098]

    = 1.0432*10^-8 i (equation 1)

    i(t)=3.67 + 1.12t²

    When t=2.00s, i = 3.67 + 1.12(2.00)² = 8.15A

    From equation 1:

    Φ = 1.0432*10^-8 * 8.15 = 8.502*10^-8 Wb

    Applying the right hand grip rule to the wire, the direction of B (hence direction of Φ) is into the page. So the flux is positive and to 3 sig. figs. is 8.50*10^-8 Wb.

    ____________________

    B. The magnitude of the induced emf is |emf| = |dΦ/dt|.

    From equation 1:

    dΦ/dt = 1.0432*10^-8 di/dt

    di/dt = 2.24t. When t=2.00s, di/dt = 2.24*2.00 = 4.48 A/s

    |emf| = 1.0432 * 10^-8 * 4.48 = 4.67*10^-8 V

    The flux is into the page and increasing (as i increases with time).  So the induced current produces an opposing field directed out of the page.  Applying right hand rule, current in loop is anticlockwise. So emf is negative, i.e. -4.67*10^-8 V

    Arithmetic no guaranteed though.

  • 5 months ago

    Kindly attach the diagram

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