Question

The magnetic induction at a point P which is at the distance of 4 cm from a long current carrying wire is $ 10^{-3}$ T. The field of induction at a distance 12 cm from the current will be

• A. $3.33 \times {10}^{-4} \,T $
• B. $1.11 \times {10}^{-4} \,T $
• C. $33 \times {10}^{-3}\, T $
• D. $ 9 \times {10}^{-3}\, T $

Answer 1

The Correct Option is A

To find the magnetic induction at a point at a distance of 12 cm from a long current-carrying wire, we use the formula for the magnetic field due to a long straight current-carrying wire:

B = \frac{\mu_0 I}{2 \pi r}

where:

• B is the magnetic induction (magnetic field) at distance r.
• \mu_0 is the permeability of free space.
• I is the current flowing through the wire.
• r is the distance from the wire where the field is being measured.

Given:

• B_1 = 10^{-3}\, T at r_1 = 4\, cm = 0.04\, m
• We need to find B_2 at r_2 = 12\, cm = 0.12\, m

We know that the magnetic field is inversely proportional to the distance from the wire:

B \propto \frac{1}{r}

Thus, we can write:

\frac{B_1}{B_2} = \frac{r_2}{r_1}

Substituting the given values:

\frac{10^{-3}}{B_2} = \frac{0.12}{0.04}

Simplifying:

\frac{10^{-3}}{B_2} = 3

Thus,

B_2 = \frac{10^{-3}}{3} = 3.33 \times 10^{-4}\, T

Therefore, the magnetic induction at a distance of 12 cm from the wire is 3.33 \times 10^{-4}\, T, which corresponds to the first option given:

• 3.33 \times 10^{-4}\, T

Hence, the correct answer is 3.33 \times 10^{-4}\, T.