A long solenoid with 70.0 turns/cm and aradius of 9.00 cm carries a current of14

Question

A long solenoid with 70.0 turns/cm and aradius of 9.00 cm carries a current of14.0 mA. A current of 8.00 A exists in a straight conductor located alongthe axis of the solenoid. (a) At what radial distance from the axis willthe direction of the resulting magnetic field be at 45.0° tothe axial direction?
1 cm

(b) What is the magnitude of the magnetic field there?
2 T
(a) At what radial distance from the axis willthe direction of the resulting magnetic field be at 45.0° tothe axial direction?
1 cm

(b) What is the magnitude of the magnetic field there?
2 T

Explanation / Answer

(a)the resulting magnetic field is B = (Bx^2 + By^2)^(1/2) B1x = B1 * cos1 B2x = B2 * cos2 B1y = B1 * sin1 B2y = B2 * sin2 Bx = B1x + B2x By = B1y + B2y here,1 = 2 = 90o therefore Bx = 0 or By = B1 + B2 B1 = (on * I1) B2 = (oI2/2a) or B = B1 + B2 -----------(1) B1 = B2 * sin or (on * I1) =(oI2/2a) * sin(45.0o) or a = (I2/nI1) * (1/2) * sin(45.0o) (b)the magnitude of the magnetic field there is obtained fromequation (1) therefore Bx = 0 or By = B1 + B2 B1 = (on * I1) B2 = (oI2/2a) or B = B1 + B2 -----------(1) B1 = B2 * sin or (on * I1) =(oI2/2a) * sin(45.0o) or a = (I2/nI1) * (1/2) * sin(45.0o) (b)the magnitude of the magnetic field there is obtained fromequation (1)

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