A steady current $I$ flows down a hollow cylindrical tube of radius a and is uniformly distributed around the tube. Let $r$ be the distance from the axis of symmetry of the tube to a given point. What is the magnitude of the magnetic field $B,$ at a point inside the tube?

z e r o

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\frac{2 I}{r c}

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\frac{2 I r}{a^{2} c}

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\frac{4 π (r - a) I}{r^{2}}

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Solution

The correct option is A $z e r o$
Consider a loop inside the hollow tube as current passing through any loop inside the tube is zero.
Hence using Ampere's circuital law
$\oint \to B d \to l = 0$ or $B = 0$

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A steady current I flows down a hollow cylindrical tube of radius a and is uniformly distributed around the tube. Let r be the distance from the axis of symmetry of the tube to a given point. What is the magnitude of the magnetic fieldB, at a point inside the tube?

The correct option is A zeroConsider a loop inside the hollow tube as current passing through any loop inside the tube is zero. Hence using Ampere's circuital law ∮→Bd→l=0 or B=0

A steady current $I$ flows down a hollow cylindrical tube of radius a and is uniformly distributed around the tube. Let $r$ be the distance from the axis of symmetry of the tube to a given point. What is the magnitude of the magnetic field $B,$ at a point inside the tube?

The correct option is A $z e r o$
Consider a loop inside the hollow tube as current passing through any loop inside the tube is zero.
Hence using Ampere's circuital law
$\oint \to B d \to l = 0$ or $B = 0$