An infinite straight wire carries current I 1 = 2.7 A in the positive y-directio
ID: 3901025 • Letter: A
Question
An infinite straight wire carries current I1= 2.7 A in the positive y-direction as shown. At time t = 0, a conducting wire, aligned with the y-direction is located a distance d = 56 cm from the y-axis and moves with velocity v = 16 cm/s in the negaitve x-direction as shown. The wire has length W = 20 cm.
1)What is ?(0), the emf induced in the moving wire at t = 0? Define the emf to be positive if the potential at point a is higher than that at point b.
2)What is ?(t1), the emf induced in the moving wire at t = t1 = 2.3 s? Define the emf to be positive if the potential at point a is higher than that at point b.
3)The wire is now replaced by a conducting rectangular loop as shown. The loop has length L = 45 cm and width W = 20 cm. At time t = 0, the loop moves with velocity v = 16 cm/s with its left end located a distance d = 56 cm from the y-axis. The resistance of the loop is R = 1.1 ?. What is i(0), the induced current in the loop at time t = 0? Define the current to be positive if it flows in the counter-clockwise direction.
4)Suppose the loop now moves in the positive y-direction as shown. What is the direction of the induced current now?
a)The current flows counterclockwise
b)The current flows clockwise
c)There is no induced current now
5)Suppose now that the loop is rotated 90o and moves with velocity v = 16 cm/s in the positive x-direction as shown. What is I2, the current in the infinite wire, if the induced current in the loop at the instant shown (d = 56 cm) is the same as it was in the third part of this problem (i.e., when the left end of loop was at a distance d = 56 cm from the y-axis)?
Please give answers, not just equations.
Explanation / Answer
Solution for same question with I1 = 4.7 A, d = 43cm, v = 15cm/s, W = 16cm, L = 57cm, R= 1.9 ohm
Please change the parameter and you would get the results.
1) Magnetic field = mu_o * I / 2*pi *d = 1.256 * 10^-6 * 4.7/ 2 * 3.14 * 0.43 = 2.18 * 10^ -6 T
Emf induced = v( B x L ) = 0.15 * ( 2.18 * 10^ -6 * 0.16) = 5.23 * 10^ -8 V
Potential at a is higher.
2) At t = 1.9 s,
Distance between two wires = 0.43 - (0.15*1.9) = 0.145m
Magnetic field = mu_o * I / 2*pi *d = 1.256 * 10^-6 * 4.7/ 2 * 3.14 * 0.145 = 6.46 * 10^ -6 T
Emf induced = v( B x L ) = 0.15 * ( 6.46 * 10^ -6 * 0.16) = 15.49 * 10^ - 8 V
Potential at a is higher.
3) Emf induced in left side of loop = 5.23 * 10^ -8 V (in clockwise)
For right side
Magnetic field = mu_o * I / 2*pi *d = 1.256 * 10^-6 * 4.7/ 2 * 3.14 * (0.57 +0.43) = 0.9374 * 10^ -6 T
Emf induced = v( B x L ) = 0.15 * ( 0.937 * 10^ -6 * 0.16) = 2.24 * 10^ -8 V ( in anti clockwise)
Net emf = 5.23 * 10^ -8 V - 2.24 * 10^ -8 = 2.99 * 10^ -8 V ( in clockwise )
I = V/r = 2.99 * 10^ -8/1.9 = - 1.57 * 10^ -8 A
4) No induced current
5) Magnetic field = mu_o * I / 2*pi *d = 1.256 * 10^-6 * 4.7/ 2 * 3.14 * 0.43 = 2.18 * 10^ -6 T
Emf induced = v( B x L ) = 0.15 * ( 2.18 * 10^ -6 * 0.57) = 18.63 * 10^ -8 V ( in clockwise)
Magnetic field = mu_o * I / 2*pi *d = 1.256 * 10^-6 * 4.7/ 2 * 3.14 * (0.43+0.16) = 1.588 * 10^ -6 T
Emf induced = v( B x L ) = 0.15 * ( 1.588 * 10^ -6 * 0.57) = 13.57 * 10^ -8 V (anti clockwise)
Net emf = 18.63 * 10^ -8 V - 13.57 * 10^ -8 = 5.06 * 10^ -8 V ( in clockwise )
I = V/r = 5.06 * 10^ -8/1.9 = - 2.66 * 10^ -8 A