QUESTION 5 What is the magnitude of the magnetic field at point P if a = R and b = 2R?
The Correct Answer and Explanation is:
To solve this problem, we’ll need to understand the context of the magnetic field. I’ll assume that the magnetic field at point P is due to a current-carrying conductor, with a certain configuration (like a circular loop, for example), and that aaa and bbb represent the distances or radii in the setup.
Let’s say the magnetic field at point P is created by a current III passing through a circular loop. If a=Ra = Ra=R and b=2Rb = 2Rb=2R, and point P lies at some position in relation to the loop, we would use the Biot-Savart law or Ampere’s Law depending on the exact configuration.
Explanation using Ampere’s Law for a Circular Loop:
For a current-carrying wire in the shape of a circle, the magnetic field at a point on the axis of the loop can be calculated using Ampere’s Law. The formula for the magnetic field at a distance xxx from the center of a circular current loop along its axis is given by:B=μ0IR22(R2+x2)3/2B = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}}B=2(R2+x2)3/2μ0IR2
Where:
- BBB is the magnetic field at point P.
- μ0\mu_0μ0 is the permeability of free space (4π×10−7 T\cdotpm/A4\pi \times 10^{-7} \, \text{T·m/A}4π×10−7T\cdotpm/A).
- III is the current through the loop.
- RRR is the radius of the loop.
- xxx is the distance from the center of the loop along the axis (where point P is located).
Case 1: a=Ra = Ra=R
If a=Ra = Ra=R, point P is at a distance equal to the radius of the loop. Plugging this into the formula:B=μ0IR22(R2+R2)3/2=μ0IR22(2R2)3/2=μ0I42RB = \frac{\mu_0 I R^2}{2 (R^2 + R^2)^{3/2}} = \frac{\mu_0 I R^2}{2 (2R^2)^{3/2}} = \frac{\mu_0 I}{4 \sqrt{2} R}B=2(R2+R2)3/2μ0IR2=2(2R2)3/2μ0IR2=42Rμ0I
This gives the magnetic field at the point P when a=Ra = Ra=R.
Case 2: b=2Rb = 2Rb=2R
For b=2Rb = 2Rb=2R, point P is at a distance of 2R2R2R along the axis of the loop. Using the formula again:B=μ0IR22(R2+(2R)2)3/2=μ0IR22(5R2)3/2=μ0I105RB = \frac{\mu_0 I R^2}{2 (R^2 + (2R)^2)^{3/2}} = \frac{\mu_0 I R^2}{2 (5R^2)^{3/2}} = \frac{\mu_0 I}{10 \sqrt{5} R}B=2(R2+(2R)2)3/2μ0IR2=2(5R2)3/2μ0IR2=105Rμ0I
This gives the magnetic field at point P when b=2Rb = 2Rb=2R.
Conclusion:
The magnetic field strength at point P will depend on the specific distance from the loop. By comparing the two cases, the field at a=Ra = Ra=R is stronger than at b=2Rb = 2Rb=2R. The closer the point is to the center of the loop (i.e., smaller RRR), the stronger the magnetic field.
