AIIMS2019Physics-Electromagnetism

AIIMS 2019 Physics Magnetic Field in Solenoids MCQ Question

Type: MCQ-conceptual-Medium-Class 12

A toroid having average diameter 2.5 m, number A turns 400, current = 2A and magnetic field has 10 T what will be induced magnetic field (in amp/m)

A

1054π\frac{10^5}{4\pi}

B

1084π\frac{10^8}{4\pi}

C

1082π\frac{10^8}{2\pi}

D

1022π\frac{10^2}{2\pi}

Correct Answer

Option B

Detailed Explanation

Option B is correct because it accurately represents the formula for calculating the induced magnetic field (B) around a current-carrying conductor, incorporating both the contributions from the magnetic field intensity (H) and the current (i) through the relationship B = μ₀(H + I), where μ₀ is the permeability of free space. Option A is incorrect as it misrepresents the relationship by adding current directly to the magnetic field intensity without proper context. Options C and D do not provide a valid expression for calculating the induced magnetic field; instead, they suggest substitution or equate it to a constant, which is not relevant to the formula needed for this calculation. Understanding these relationships is crucial for analyzing magnetic fields in electromagnetism.

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