91Ó°ÊÓ

Magnetic flux is given by

$\mathit{ϕ}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{A}}$

Where$\stackrel{\mathbf{→}}{\mathbf{B}}$ is the magnetic field and$\stackrel{\mathbf{→}}{\mathbf{A}}$is the area vector

Faraday’s law states that The induced emf in a coil is equal to the negative of the rate of change of magnetic flux times the number of turns in the coil. It involves the interaction of charge with magnetic field.

$\mathit{e}\mathit{m}\mathit{f}\mathbf{=}\mathbf{-}\mathit{N}\frac{\mathbf{∆}\mathbf{ϕ}}{\mathbf{∆}\mathbf{t}}$

Replacing the change of flux into faraday’s law we get

$\mathrm{emf}=-\mathrm{N}\frac{\left(\mathrm{BA}\right)}{∆\mathrm{t}}$

Replace N=1

And since the area does not change and only magnetic field changes.

$\mathrm{Î\mu }=\mathrm{A}\frac{\mathrm{dB}}{\mathrm{dt}}$

The term$\frac{\mathrm{dB}}{\mathrm{dt}}$ is the slope of the curve, Hence the steeper is the curve the greater the produced emf and so the induced current.

The Magnetic field changes from 0 to 4.0T in about 0.15s.Thus

$\frac{∆\mathrm{B}}{∆\mathrm{t}}=\frac{4.0\mathrm{T}}{0.15\mathrm{ms}}=16.7×{10}^3\mathrm{T}/\mathrm{s}$

In the graph A, the magnetic field changes from 0 to 4.0T in about 0.10s

$\frac{∆\mathrm{B}}{∆\mathrm{t}}=\frac{4.0\mathrm{T}}{0.10\mathrm{ms}}=40.0×{10}^3\mathrm{T}/\mathrm{s}$

In graph B, the magnetic field changes from 0 to 4.0T in about 0.10s

$\frac{∆\mathrm{B}}{∆\mathrm{t}}=\frac{4.0\mathrm{T}}{0.10\mathrm{ms}}=40.0×{10}^3\mathrm{T}/\mathrm{s}$

So both the graphs A and B would achieve the goal of increasing the maximum induced current in the brain

Hence Option C is the right choice.