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Question

Physics Question on Moving charges and magnetism

A long straight wire carrying a current of 30 A is placed in an external uniform magnetic field. of induction 4×10−44\times {{10}^{-4}} T. The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesia at a point 2.0 cm away from the wire is (μ0=4π×10−7H/m)({{\mu }_{0}}=4\pi \times {{10}^{-7}}H/m)

A

10−4{{10}^{-4}}

B

3×10−43\times {{10}^{-4}}

C

5×10−45\times {{10}^{-4}}

D

6×10−46\times {{10}^{-4}}

Answer

5×10−45\times {{10}^{-4}}

Explanation

Solution

Magnetic field induction at point P due to current carrying wire is B2=μ0i2πr4π×10−7×302π×0.02{{B}_{2}}=\frac{{{\mu }_{0}}i}{2\pi r}\frac{4\pi \times {{10}^{-7}}\times 30}{2\pi \times 0.02} =(4)2+(2)2×10−7=\sqrt{{{(4)}^{2}}+{{(2)}^{2}}\times {{10}^{-7}}} B2=3×10−4T{{B}_{2}}=3\times {{10}^{-4}}T The direction of B2B_2 will be perpendecular to S, then, the magnitude of the resultant magnetic induction B=B12+B22B=\sqrt{B_{1}^{2}+B_{2}^{2}} =(4)2+(3)2×10−4=\sqrt{{{(4)}^{2}}+{{(3)}^{2}}\times {{10}^{-4}}} =5×10−4T=5\times {{10}^{-4}}T