Three long, parallel conductors each carry a current of/= 3.65 A. Figure P30.15

Question

Three long, parallel conductors each carry a current of/= 3.65 A. Figure P30.15 is an end view of the conductors, with each current coming out of the page. Taking a = 1.21 cm, determine the magnitude and direction of the magnetic field at point A. N.B. A positive response indicates a magnetic field directed towards the top of the page, while a negative response indicates a magnetic field directed towards the bottom of the page. Determine the magnitude and direction of the magnetic field at point B. Determine the magnitude and direction of the magnetic field at point C.

Explanation / Answer

a) field due to current carrying wire at distance d is

B = u0 I / (2 pi d )

AT A:

due to upper wire will be 45 deg below.

magnitude B1 = [(4pi x 10^-7 x 3.65 ) / (2 x pi x 0.0121) ] (cos45i - sin45j]

B1 =( 4.266i - 4.266j ) x 10^-7 T

due to lower wire:

B2 = [(4pi x 10^-7 x 3.65 ) / (2 x pi x 0.0121) ] (- cos45i - sin45j]

B2 =( - 4.266i - 4.266j ) x 10^-7 T

due to wire to the right:

B3 = [(4pi x 10^-7 x 3.65 ) / (2 x pi x 3 x 0.0121) ](-j)

    = - 2.011 x 10^-5 j

Bnet = B1 + B2 + B3 = - 10.543 x 10^-5 T

B) at B:

net field due to upper and lower wire will be zero.

due to wire to the right.

B = [(4pi x 10^-7 x 3.65 ) / (2 x pi x 2 x 0.0121) ](-j)

    = - 2.016 x 10^-5 T

C)
AT C:

magnitude B1 = [(4pi x 10^-7 x 3.65 ) / (2 x pi x 0.0121) ] (cos45i + sin45j]

B1 =( 4.266i + 4.266j ) x 10^-7 T

due to lower wire:

B2 = [(4pi x 10^-7 x 3.65 ) / (2 x pi x 0.0121) ] (- cos45i + sin45j]

B2 =( - 4.266i + 4.266j ) x 10^-7 T

due to wire to the right:

B3 = [(4pi x 10^-7 x 3.65 ) / (2 x pi x 3 x 0.0121) ](-j)

    = - 2.011 x 10^-5 j

Bnet = B1 + B2 + B3 = 6.441 x 10^-5 T

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