Question

The magnetic flux density B is changing in magnitude at a constant rate $\dfrac{{dB}}{{dt.}}$. 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 magnetic field B. The induced current in the loop is $i$. The resistivity of copper is $\rho $and density is $d$.The value of the induced current $i$is
A) $\dfrac{m}{{2\pi \rho d}}\dfrac{{dB}}{{dt}}$
B) $\dfrac{m}{{2\pi {a^2}d}}\dfrac{{dB}}{{dt}}$
C) $\dfrac{m}{{4\pi ad}}\dfrac{{dB}}{{dt}}$
D) $\dfrac{m}{{4\pi \rho d}}\dfrac{{dB}}{{dt}}$

Answer 1

Hint:By Faraday’s law, the induced emf is defined as the rate of change of flux, therefore the induced current is given by,
$i = \dfrac{\varepsilon }{R} = \dfrac{1}{R}\dfrac{{d\phi }}{{dt}}$
$\therefore i = \dfrac{1}{R}\dfrac{{d\left( {BA} \right)}}{{dt}}$
According to question the area of the loop is equal to $\pi {r^2}$ and the area of the cross section of wire will be $\pi {a^2}$.With the help of above relation we can easily calculate the induced current in the copper wire.

Complete step-by-step answer:
We know that in electromagnetic induction the induced current is given by the formula
$i = \dfrac{1}{R}\dfrac{{d\left( {BA} \right)}}{{dt}}$
$i = \dfrac{A}{R}\dfrac{{dB}}{{dt}}$
Where
i = induced current
A = area of loop perpendicular to magnetic field
R = Resistance of copper wire
According to question
Length of wire = circumference of loop
$l = 2\pi r \Leftrightarrow r = \dfrac{l}{{2\pi }} \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)$
Let the Area of the loop be$A = \pi {r^2}$
Let the Area of cross section of wire be ${A'} = \pi {a^2}$
Therefore the Resistance of the wire given by the equation
$R = \rho \dfrac{l}{{{A'}}}$
In order calculate first we need to calculate the value of $\dfrac{A}{R}$
$\therefore \dfrac{A}{R} = \dfrac{{\pi {r^2}}}{\rho } \times \dfrac{{\pi {a^2}}}{l} \cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right)$
From (1) and (2) we get
$\therefore \dfrac{A}{R} = \dfrac{{\pi {{\left( {\dfrac{l}{{2\pi }}} \right)}^2}}}{\rho } \times \dfrac{{\pi {a^2}}}{l}$
On further solving we get
$\therefore \dfrac{A}{R} = \dfrac{{l{a^2}}}{{4\rho }} \cdot \cdot \cdot \cdot \cdot \cdot \left( 3 \right)$
Also we know that,
$mass = volume \times density$
$m = ({A'} \times l) \times d \Leftrightarrow l = \dfrac{m}{{\pi {a^2} \times d}} \cdot \cdot \cdot \cdot \cdot \cdot (4)$
From (3) and (4) we get
$\therefore \dfrac{A}{R} = \dfrac{{\left( {\dfrac{m}{{\pi {a^2}d}}} \right){a^2}}}{{4\rho }} \Leftrightarrow \dfrac{A}{R} = \dfrac{m}{{4\pi \rho d}} \cdot \cdot \cdot \cdot \cdot \cdot \left( 5 \right)$
Putting the final value of $\dfrac{A}{R} = \dfrac{m}{{4\pi \rho d}}$ on the formula of induced current we get
$i = \dfrac{A}{R}\dfrac{{dB}}{{dt}}$
$\therefore i = \dfrac{m}{{4\pi \rho d}}\dfrac{{dB}}{{dt}}$
Hence the correct option is D.

Note:
In the above question, we have only considered the magnitude of the induced current. By Faraday’s law of electromagnetic induction, an induced current has a direction such that the magnetic field due to induced current opposes the change in the magnetic flux that induces the current. An induced current and emf opposes the change.