QJEEM25Phf88f: JEE Main 2025 Physics - Class 12 JEEM25Phf88f Solution: Consider a long straight wire of a circular cross-section…

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QJEEM25Phf88f: JEE Main 2025 Physics - Class 12 JEEM25Phf88f Solution: Consider a long straight wire of a circular cross-section…

Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire’s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be :

Multiple Choice Options

A [a/2, 3a]
B [a/4, 3a/2]
C [a/2, 2a]
D [a/4, 2a]

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Accepted Answer

Maximum possible magnetic field is at the surface

$\begin{aligned} B_{max} = \frac{μ_{0} I}{2 π a} \\ \frac{B_{max}}{2} = \frac{μ_{0} I}{4 π a} \end{aligned}$

It can be obtained inside as well as outside the wire

For inside,

$\begin{aligned} \frac{μ_{0} I}{4 π a} = \frac{μ_{0} Ir}{2 π a^{2}} \\ \Rightarrow r = \frac{a}{2} \end{aligned}$

For outside

$\begin{aligned} \frac{μ_{0} I}{4 π a} = \frac{μ_{0} I}{2 π r} \\ \Rightarrow r = 2 a \end{aligned}$

Correct answer $\left[\frac{\mathrm{a}}{2}, 2 \mathrm{a}\right]$

QJEEM25Phf88f: JEE Main 2025 Physics - Class 12 JEEM25Phf88f Solution: Consider a long straight wire of a circular cross-section…

Multiple Choice Options

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