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Magnetic induction at a point on the surface of a long straight current carrying wire is 12 $μ T$ . Then magnetic induction at a distance 3R from its surface is (R = radius of the wire)

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6 $μ T$

3 $μ T$

4 $μ T$

8 $μ T$

answer is B.

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Detailed Solution

$\frac{μ_{0} i}{2 π R} = 12 μ T$

At a distance 3R from the surface, $B = \frac{μ_{0} i}{2 π (R + 3 R)} = \frac{1}{4} \times \frac{μ_{0} i}{2 π R} = \frac{12}{4} μ T$

$\Rightarrow B = 3 μ T$

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