Question:
A long current carrying wire produces a magnetic field of \(1\ \text{T}\) at a distance \(r\). The magnetic field at
\[
(a)\ \frac{r}{2}, \qquad (b)\ 2r, \qquad (c)\ 3r
\]
is
A long current carrying wire produces a magnetic field of \(1\ \text{T}\) at a distance \(r\). The magnetic field at
\[ (a)\ \frac{r}{2}, \qquad (b)\ 2r, \qquad (c)\ 3r \] is
\[ (a)\ \frac{r}{2}, \qquad (b)\ 2r, \qquad (c)\ 3r \] is
Show Hint
For a long straight current carrying wire:
\[
B=\frac{\mu_0 I}{2\pi r}
\]
The magnetic field decreases inversely with distance from the wire.
Updated On: Jun 25, 2026
- \((a)\ 2\text{T},\ (b)\ \dfrac{1}{2}\text{T},\ (c)\ \dfrac{1}{3}\text{T}\)
- \((a)\ 3\text{T},\ (b)\ \dfrac{1}{3}\text{T},\ (c)\ \dfrac{1}{6}\text{T}\)
- \((a)\ \dfrac{3}{2}\text{T},\ (b)\ \dfrac{1}{4}\text{T},\ (c)\ \dfrac{1}{8}\text{T}\)
- \((a)\ \dfrac{5}{2}\text{T},\ (b)\ \dfrac{1}{2}\text{T},\ (c)\ \dfrac{1}{3}\text{T}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Write the formula for magnetic field due to a long straight current carrying wire.
The magnetic field at a distance \(r\) from a long straight wire is \[ B=\frac{\mu_0 I}{2\pi r} \] Thus, \[ B\propto \frac{1}{r} \] So, magnetic field is inversely proportional to distance from the wire.
Given: \[ B=1\ \text{T} \] at distance \[ r \]
Step 2: Find magnetic field at \(\dfrac{r}{2}\).
Since \[ B\propto \frac{1}{r}, \] halving the distance doubles the magnetic field.
Therefore, \[ B_{r/2}=2\times 1 \] \[ B_{r/2}=2\ \text{T} \]
Step 3: Find magnetic field at \(2r\).
Doubling the distance halves the magnetic field.
Hence, \[ B_{2r}=\frac{1}{2}\times 1 \] \[ B_{2r}=\frac{1}{2}\ \text{T} \]
Step 4: Find magnetic field at \(3r\).
Tripling the distance reduces the magnetic field to one-third.
Thus, \[ B_{3r}=\frac{1}{3}\times 1 \] \[ B_{3r}=\frac{1}{3}\ \text{T} \]
Step 5: Final conclusion.
Therefore, \[ (a)\ 2\text{T},\qquad (b)\ \frac{1}{2}\text{T},\qquad (c)\ \frac{1}{3}\text{T} \] Hence, the correct option is \[ \boxed{(1)} \]
The magnetic field at a distance \(r\) from a long straight wire is \[ B=\frac{\mu_0 I}{2\pi r} \] Thus, \[ B\propto \frac{1}{r} \] So, magnetic field is inversely proportional to distance from the wire.
Given: \[ B=1\ \text{T} \] at distance \[ r \]
Step 2: Find magnetic field at \(\dfrac{r}{2}\).
Since \[ B\propto \frac{1}{r}, \] halving the distance doubles the magnetic field.
Therefore, \[ B_{r/2}=2\times 1 \] \[ B_{r/2}=2\ \text{T} \]
Step 3: Find magnetic field at \(2r\).
Doubling the distance halves the magnetic field.
Hence, \[ B_{2r}=\frac{1}{2}\times 1 \] \[ B_{2r}=\frac{1}{2}\ \text{T} \]
Step 4: Find magnetic field at \(3r\).
Tripling the distance reduces the magnetic field to one-third.
Thus, \[ B_{3r}=\frac{1}{3}\times 1 \] \[ B_{3r}=\frac{1}{3}\ \text{T} \]
Step 5: Final conclusion.
Therefore, \[ (a)\ 2\text{T},\qquad (b)\ \frac{1}{2}\text{T},\qquad (c)\ \frac{1}{3}\text{T} \] Hence, the correct option is \[ \boxed{(1)} \]
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