The magnetic flux density B is changing in magnitude at a constant rate dB/dt. A given mass m of copper, drawn into a wire of radius a and formed into a circular loop of radius r is placed perpendicular to the field B. The induced current in the loop is i. The resistivity of copper is $ρ$ and density is d. The value of the induced current i is

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$\frac{m}{2 π ρ d} \frac{d B}{d t}$

$\frac{m}{4 π a^{2} r} \frac{d B}{d t}$

$\frac{m}{4 π a d} \frac{d B}{d t}$

$\frac{m}{4 π ρ d} \frac{d B}{d t}$

answer is D.

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Detailed Solution

$e = \frac{d ϕ}{d t} \Rightarrow i = \frac{e}{R} = \frac{1}{R} \frac{d}{d t} (B A) = \frac{A}{R} \frac{d B}{d t}$

where $π r^{2} =$ area of the loop of radius r and R = resistance of the loop of length $(2 π r)$ and area of cross section $π a^{2}$ .

$R = \frac{ρ l}{π a^{2}} = \frac{ρ (2 π r)}{π a^{2}}$

Further mass of wire is $m = (π a^{2}) (2 π r) (d)$

$i = \frac{(π a^{2}) (π r^{2})}{ρ (2 π r)} \frac{d B}{d t}$

$i = \frac{(π a^{2}) (2 π r)}{4 π ρ} \frac{d B}{d t} \Rightarrow i = \frac{m}{4 π ρ d} \frac{d B}{d t}$

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