A long, straight, solid cylindrical conductor of radius R carries a steady current I that is uniformly distributed over its cross-section. As shown in the figure, point P is located at a radial distance of R 4 from the axis, and point Q is located at a radial distance of 2R from the axis. If B_P and B_Q are the magnitudes of the magnetic fields at points P and Q respectively, what is the ratio B_P B_Q ?

Step-by-step solution

For a solid cylindrical conductor of radius R carrying a uniformly distributed current I , the magnetic field at a distance r from the axis is given by Ampere's law: Inside the conductor ( r B_ in = ₀ I r 2 R^2 Outside the conductor ( r > R ): B_ out = ₀ I 2 r For point P , the distance from the axis is r_P = R 4 (which is less than R ). Thus, the magnetic field at P is: B_P = ₀ I ( R 4 ) 2 R^2 = ₀ I 8 R For point Q , the distance from the axis is r_Q = 2R (which is greater than R ). Thus, the magnetic field at Q i