the magnetic field around a current - carrying wire is proportional to the current and proportional to the…

1. Recall the formula for the magnetic - field strength around a current - carrying wire:
   • The magnetic - field strength (B) around a long straight current - carrying wire is given by (B=\frac{\mu_0I}{2\pi r}), where (\mu_0) is the permeability of free space, (I) is the current in the wire, and (r) is the distance from the wire. This shows that (B) is directly proportional to the current (I) and inversely proportional to the distance (r) from the wire.
2. Let the initial current be (I_1) and the initial distance be (r_1), so the initial magnetic - field strength (B_1=\frac{\mu_0I_1}{2\pi r_1}).
   • The new current (I_2 = 3I_1) (since the current triples) and the new distance (r_2=2r_1) (since the distance doubles).
   • The new magnetic - field strength (B_2=\frac{\mu_0I_2}{2\pi r_2}=\frac{\mu_0(3I_1)}{2\pi(2r_1)}).
3. Calculate the ratio (\frac{B_2}{B_1}):
   • Substitute the expressions for (B_1) and (B_2) into the ratio (\frac{B_2}{B_1}).
   • (\frac{B_2}{B_1}=\frac{\frac{\mu_0(3I_1)}{2\pi(2r_1)}}{\frac{\mu_0I_1}{2\pi r_1}}).
   • The (\mu_0), (2\pi), (I_1), and (r_1) terms cancel out. We get (\frac{B_2}{B_1}=\frac{3}{2}=1.5).

Explanation:

Step1: Identify the magnetic - field formula

(B = \frac{\mu_0I}{2\pi r}), showing (B\propto I) and (B\propto\frac{1}{r})

Step2: Define initial and new values

Let (I_1) and (r_1) be initial values, (I_2 = 3I_1), (r_2 = 2r_1)

Step3: Calculate new magnetic - field (B_2)

(B_2=\frac{\mu_0I_2}{2\pi r_2}=\frac{\mu_0(3I_1)}{2\pi(2r_1)})

Step4: Find the ratio (\frac{B_2}{B_1})

(\frac{B_2}{B_1}=\frac{\frac{\mu_0(3I_1)}{2\pi(2r_1)}}{\frac{\mu_0I_1}{2\pi r_1}}=\frac{3}{2})

Answer:

1.5