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Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 27 January, Shift 1 — Question 55

Two long, straight wires carry equal currents in opposite directions as shown in figure. The separation between the wires is 5.0 cm . The magnitude of the magnetic field at a point PP midway between the wires is _______\_\_\_\_\_\_\_ μT\mu \mathrm{T}

(Given : μ0=4π×10−7TmA−1\mu_{0}=4 \pi \times 10^{-7} \mathrm{TmA}^{-1} ) 10 A∣P˙∣10 A10 \mathrm{~A}|\dot{\mathrm{P}}| 10 \mathrm{~A}

Question figure

Answer: 160

Numerical answer — enter this value.

Step-by-step solution

B=(μ0i2πa)×2=4π×10−7×10π×(52×10−2)B=\left(\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{a}}\right) \times 2=\frac{4 \pi \times 10^{-7} \times 10}{\pi \times\left(\frac{5}{2} \times 10^{-2}\right)}

=16×10−5=160μ T=16 \times 10^{-5}=160 \mu \mathrm{~T}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 1
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law

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