# The component of the external magnetic field along the central axis of a 125-turn circular coil...

## Question:

The component of the external magnetic field along the central axis of a 125-turn circular coil of radius 18.0 cm decreases from 1.90 T to 0.350 T in 2.40 s. If the resistance of the coil is 6.00 {eq}\Omega {/eq}, what is the magnitude of the induced current in the coil? What is the direction of the current if the axial component of the field points away from the viewer? Clockwise or Counterclockwise?

## Faraday's Law of Electromagnetic Induction:

Faraday's law says that EMF can be induced in a coil, if magnetic field through the coil is changed in magnitude or direction, w.r.t. the time. The magnitude of induced EMF can be calculated using the Faraday's law. The direction of the induced current can be find out by the help of Lenz's law. It is given by the expression:

{eq}\begin{align} \epsilon & = -\dfrac{\Delta \phi}{\Delta t} \\ \epsilon & = - A \dfrac{\Delta B }{\Delta t} \\ \end{align} {/eq}.

where,

• A is the area of the coil, which is {eq}A = \pi r^2 {/eq}
• {eq}\Delta B {/eq} is the change in magnetic field a solenoid, and
• {eq}\Delta t {/eq} is time.

## Answer and Explanation:

Given:

• Number of turns in a circular coil is {eq}N = 125 \ \text {turns} {/eq}
• The radius of a coil is {eq}r = 18.0 \ cm = 18.0 \times 10^{-2} \ m {/eq}
• The change in magnetic field is {eq}dB = B_2 - B_1 = 0.350 - 1.90 = -1.55 \ T {/eq}
• The angle between the magnetic field lines and the normal (perpendicular) to A is {eq}\theta = 0^\circ {/eq}
• The time duration is {eq}dt = 2.40 \ s {/eq}
• The resistance of the coil is {eq}R = 6.00 \ \Omega {/eq}

Let

• The magnitude of the induced current in the coil is {eq}I {/eq}.
• The magnitude of the average EMF induced in the coil is {eq}\varepsilon {/eq}.

After plugging the given values in the formula of average induced emf, we get the magnitude of induced current in the coil ;

{eq}\begin{align} \varepsilon & = - N \dfrac { d\phi} {dt} \\ IR & = - N \dfrac { d } {dt} (BA\cos\theta ) \\ 6.00 I & = - N A \cos\theta \dfrac { dB } {dt} \\ 6.00 I & = - N A \cos\theta \dfrac { (B_2 -B_1 ) } {dt} \\ 6.00 I & = - (125) \pi (18.0 \times 10^{-2})^2 \ cos0^\circ \left \{ \dfrac { -1.55 } { 2.40 } \right \} \\ 6.00 I & = -( 12.7 ) ( -0.65 ) \\ I & = \dfrac {8.26 } {6.00 } \\ \implies I & = 1.38 \ A \\ \end{align} {/eq}

Here, the axial component of the field points away from the viewer and the induced emf is in opposite direction, so according to the right hand rule the induced current is flowing counterclockwise.

Hence, the magnitude of the induced current in the coil is {eq}1.38 \ A {/eq} which is flowing counterclockwise.