Magnetic Effects & Magnetism

JEE Main level, single correct, medium

77 of 276

Question 77 of 276

Find the magnetic field at point PP due to a straight line segment ABAB of length 6cm6cm carrying a current of 5A.5A. (See figure) (μ0=4π×10−7NA−2)\left({\mu }_{0}=4\pi \times {10}^{-7}{NA}^{-2}\right)

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Answer: (A) 1.5×10−5T1.5\times {10}^{-5}T

θ=370\theta ={37}^{0}
r=4cmr=4cm
B=μ0i4πr[sin37∘+sin37∘]B=\dfrac{{\mu }_{0}i}{4\pi r}[sin37^{\circ}+sin37^{\circ}]
=4π×10−7(5)4π×(4×10−2)(35+35)=\dfrac{4\pi \times {10}^{-7}\left(5\right)}{4\pi \times \left(4\times {10}^{-2}\right)}\left(\dfrac{3}{5}+\dfrac{3}{5}\right)
=32×10−5=1.5×10−5T=\dfrac{3}{2}\times {10}^{-5}=1.5\times {10}^{-5}T

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