A long straight wire carrying a current of 30A is placed in an external uniform magnetic field of induction 4×10-4T. The magnetic field is acting parallel to the direction of the current. The magnitude of the resultant magnetic induction in tesla at a point 2.0cm away from the wire is : [μ0 = 4π×10-7Hm-1] (1) 10-4 (2) 3×10-4 (3) 5×10-4 (4) 6×10-4

A long straight wire carrying a current of 30A is placed in an external uniform magnetic field of induction 4×10<sup>-4</sup>T. The magnetic field is acting parallel to the direction of the current. The magnitude of the resultant magnetic induction in tesla at a point 2.0cm away from the wire is : [μ0 = 4π×10<sup>-7</sup>Hm<sup>-1</sup>] (1) 10<sup>-4</sup> (2) 3×10<sup>-4</sup> (3) 5×10<sup>-4</sup> (4) 6×10<sup>-4</sup>

Answer: 5×10-4

Current (I) = 30A

Uniform magnetic feild (B1) = 4×10-4T

r = 2 cm = 0.02 m

Magnetic field induction at point P due to current carrying wire is,

\(\begin{array}{l}B_{2}= frac{mu _{0}I}{2pi r}\end{array} \)

\(\begin{array}{l}B_{2}= frac{4pi times 10^{-7}times 30}{2pi times 0.02}\end{array} \)

B1 = 3 × 10-4T

Resultant magnetic Induction can be calculated as

\(\begin{array}{l}B = sqrt{B_{1}^{2}+B_{2}^{2}}\end{array} \)

\(\begin{array}{l}B = sqrt{(3times 10^{-4})^{2})+(4times 10^{-4})^{2}}\end{array} \)

B = 5 ×10-4T

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