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A long straight wire carrying a current of 30   A is placed in an external uniform magnetic field of induction 4 × 10 – 4   T . The direction of external magnetic field is along the direction of the current. The magnitude of the resultant magnetic induction at a point 2.0   c m away from the wire is n × 10 - 4   T . What is the value of n ?

Correct answer

5

Step-by-step solution

Magnetic field due to wire B = μ 0 I 2 π r = 4 π × 1 0 - 7 2 π × 30 2 × 1 0 - 2 = 3 × 10 - 4 T This magnetic field will be perpendicular to external magnetic field. ∴ Net magnetic field B = B 2 + B 0 2 = 3 × 1 0 – 4 2 + 4 × 1 0 – 4 2 = 5 × 10 – 4 T

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