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Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 5 April, Shift 2 — Question 50

A solenoid of length 0.5 m has a radius of 1 cm and is made up of ' mm ' number of turns. It carries a current of 5 A . If the magnitude of the magnetic field inside the solenoid is 6.28×10−3 T6.28 \times 10^{-3} \mathrm{~T}, then the value of mm is :

Answer: 500

Numerical answer — enter this value.

Step-by-step solution

μ0ni=Bn=\mu_{0} \mathrm{ni}=\mathrm{B} \quad \mathrm{n}= number of turns per unit length μ0(mℓ)i=B\mu_{0}\left(\frac{m}{\ell}\right) \mathrm{i}=\mathrm{B}

m= B. ℓμ0i=6.28×10−3×0.512.56×10−7×5\mathrm{m}=\frac{\text { B. } \ell}{\mu_{0} \mathrm{i}}=\frac{6.28 \times 10^{-3} \times 0.5}{12.56 \times 10^{-7} \times 5}

m=500\mathrm{m}=500

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
5 April, Shift 2
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Ampère's Law and its Applications

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