Answered: the current induced in the loop is zero…

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Question

The south pole of a bar magnet is moved toward a short helical coil of wire (solenoid) along the axis of the coil. The coil has 100 turns and the ends of the coil are connected to form a closed circuit. If the coil is replaced with a single loop of wire of the same diameter, and the magnet is moved exactly as before, the current induced in the loop is
zero in both cases.
100 times larger.
100 times smaller.
the same as in the coil.

Expert Solution

Step 1

Faraday's Law of Electromagnetic Induction

According to Faraday's Law of electromagnetic induction

Whenever there is a change in magnetic flux linked to a coil, an emf is induced in the coil.
The induced emf is equal to the rate of change of magnetic flux.

Lenz's Law

The direction of the induced emf is such that it opposes the cause for which it has been generated.

If $ε$ is the induced emf then

$ε = - \frac{d ϕ}{d t}$

$ϕ$ is the magnetic flux.

The magnetic flux due to a magnetic field $B$ perpendicular to an area $A$ is given by the relation

$ϕ = B A$

Step 2

Let $B$ be the magnetic field due to the bar magnet. Let us first calculate the flux through the coil.

Let $r$ be the radius of the cross-section of the coil. The flux due through one turn of the coil is

$ϕ = B A = B (π r^{2})$

Since there are 100 turns in the coil, the total flux through all the turns is

$ϕ_{1} = 100 \times B (π r^{2})$

Let the increase in flux from $0$ to $ϕ_{1}$ occur in $t$ seconds. Therefore the magnitude of the emf induced in the coil is

$ε = \frac{ϕ_{1} - 0}{t} = \frac{100 B (π r^{2})}{t}$

Let the resistance in the circuit be $R$ . Therefore the current induced is

$I_{1} = \frac{ε}{R} = \frac{100 B (π r^{2})}{R t}$

Let the coil be replaced by a single loop of the same radius, then the magnetic flux due to the magnet of the loop is

$ϕ_{2} = B (π r^{2})$

Let the increase in the magnetic flux from $0$ to $ϕ_{2}$ occur in $t$ seconds. Therefore the magnitude of the emf induced in the loop is

$ε = \frac{ϕ_{2} - 0}{t} = \frac{B (π r^{2})}{t}$

If $R$ be the resistance in the circuit then the current induced in the loop is

$I_{2} = \frac{B (π r^{2})}{R t}$

Taking the ratio

$\frac{I_{2}}{I_{1}} = \frac{\frac{B (π r^{2})}{R t}}{\frac{100 B (π r^{2})}{R t}} = \frac{1}{100}$

This shows that the current induced in the loop is 100 times smaller than that of the induced current in the coil.

Step by step

Solved in 3 steps

the current induced in the loop is zero in both cases. 100 times larger. 100 times smaller. the same as in the coil.