JEE Main 2022 · 25 Jun · Shift 1 · Physics Q40

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PhysicsSingle correct+4 / -1Question 40 of 90

A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r (r < R) from its centre will be :

B\propto r^2

B\propto r

B\propto\dfrac1{r^2}

B\propto\dfrac1r

Answer and worked solution

Official answer: B

Ampère's law inside. The enclosed current is $I\dfrac{r^2}{R^2}$ , so $B(2\pi r)=\mu_0I\dfrac{r^2}{R^2}$ , i.e. $B=\dfrac{\mu_0Ir}{2\pi R^2}\propto r$ .

(B)

Topics:Ampere's circuital law

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JEE Main 2022 · 25 Jun · Shift 1, Physics question 40