Question

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The metal equilateral triangle in the figure,

20 cm on each side, is halfway into a 0.1 T magnetic field.

A) What is the magnetic flux through the triangle?
B) If the magnetic field strength decreases, what is the direction of the induced current in the triangle?

Answer 1

Concepts and reason

The concepts required to solve the given questions is magnetic flux and direction of induced current due to change in magnetic flux that is Lentz law.

Initially, calculate the area of the triangle which is in magnetic field, later calculate the magnetic flux through the triangle by using area and magnetic field, and finally determine the direction of induced current by using Lentz law.

Fundamentals

The expression for the area of the equilateral triangle is as follows:

$A = \frac{{{a^2}\sqrt 3 }}{4}$

Here, a is length of side of the triangle.

The expression for the magnetic flux is as follows:

$\phi = NBA\cos \theta$

Here, N is the number of turns, B is the magnetic field. A is the area of coil, and $\theta$ is the angle.

The expression for the induced emf is as follows:

$\varepsilon = N\frac{{d\phi }}{{dt}}$

Here, $d\phi$ is the change in the flux and dt is the change in the time.

(A)

As half of the triangle is inside the magnetic field and half is outside the magnetic field. Thus, area of triangle $A'$which is in magnetic field is equal to half of the area of the equilateral triangle.

$A' = \frac{A}{2}$

Substitute $\frac{{{a^2}\sqrt 3 }}{4}$for A in the above equation.

$\begin{array}{c}\\A' = \frac{{\frac{{{a^2}\sqrt 3 }}{4}}}{2}\\\\ = \frac{{{a^2}\sqrt 3 }}{8}\\\end{array}$

Substitute 0.20 m for a in the equation $A' = \frac{{{a^2}\sqrt 3 }}{8}.$

$\begin{array}{c}\\A' = \frac{{{{\left( {0.20{\rm{ m}}} \right)}^2}\sqrt 3 }}{8}\\\\ = \frac{{0.04{\rm{ }}{{\rm{m}}^2}\sqrt 3 }}{8}\\\\ = 0.00866{\rm{ }}{{\rm{m}}^2}\\\end{array}$

The magnetic flux through triangle is given as follows:

$\phi = NBA\cos \theta$

Substitute $0.00866{\rm{ }}{{\rm{m}}^2}$ for A, 0.10 T for B, and $0{\rm{ }}^\circ$ for $\theta$ using the equation $\phi = NBA\cos \theta .$

$\begin{array}{c}\\\phi = \left( 1 \right)\left( {0.10{\rm{ T}}} \right)\left( {0.00866{\rm{ }}{{\rm{m}}^2}} \right)\cos 0^\circ \\\\ = \left( 1 \right)\left( {0.10{\rm{ T}}} \right)\left( {0.00866{\rm{ }}{{\rm{m}}^2}} \right)\left( 1 \right)\\\\ = 0.000866{\rm{ T}} \cdot {{\rm{m}}^2}\\\end{array}$

(B)

If the magnetic field is decreased. Then, according to the formula $\varepsilon = N\frac{{d\phi }}{{dt}}$, the value of the emf is also decreased. The decreases in the emf results in the decrease in the flux. Therefore, the induced current will be such that it will increase the flux.

Thus, the direction of the current is clockwise.

Ans: Part A

The magnetic flux through the triangle is $0.000866{\rm{ T}} \cdot {{\rm{m}}^2}$.

Part B

The direction of the induced current is clockwise direction.