Question

In: Computer Science

The metal equilateral triangle in the figure, 20 cm on each side, is halfway into a 0.1 T magnetic field.

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?

Solutions

Expert Solution

Concepts and reason

The concepts required to solve the given questions are 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 the 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 the length of the side of the triangle. The expression for the magnetic flux is as follows:

\(\phi=N B A \cos \theta\)

Here, \(\mathrm{N}\) is the number of turns, \(\mathrm{B}\) is the magnetic field. \(\mathrm{A}\) is the area of the coil, and \(\theta\) is the angle. The expression for the induced emf is as follows:

\(\varepsilon=N \frac{d \phi}{d t}\)

Here, \(d \phi\) is the change in the flux and \(\mathrm{dt}\) 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, the area of triangle \(A^{\prime}\) which is in the magnetic field is equal to half of the area of the equilateral triangle. \(A^{\prime}=\frac{A}{2}\)

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

$$ \begin{aligned} A^{\prime} &=\frac{\frac{a^{2} / 3}{4}}{2} \\ &=\frac{a^{2} \sqrt{3}}{8} \end{aligned} $$

Substitute \(0.20 \mathrm{~m}\) for \(\mathrm{a}\) in the equation \(A^{\prime}=\frac{a^{2} \sqrt{3}}{8}\).

$$ A^{\prime}=\frac{(0.20 \mathrm{~m})^{2} \sqrt{3}}{8} $$

$$ \begin{array}{l} =\frac{0.04 \mathrm{~m}^{2} \sqrt{3}}{8} \\ =0.00866 \mathrm{~m}^{2} \end{array} $$

The given triangle is an equilateral triangle because all sides are equal. The magnetic flux through a given equilateral triangle is only due to the area which is in the magnetic field. That is only due to half of the area of the triangle which is in the magnetic field.

The magnetic flux through the triangle is given as follows:

\(\phi=N B A \cos \theta\)

Substitute \(0.00866 \mathrm{~m}^{2}\) for \(\mathrm{A}, 0.10 \mathrm{~T}\) for \(\mathrm{B}\), and \(0^{\circ}\) for \(\theta\) using the equation \(\phi=N B A \cos \theta\).

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

The triangle having all the sides equal is an equilateral triangle. As half of the triangle is inside the magnetic field and half is outside the magnetic field. Thus, the area of the equilateral is half of the total area of the coil.

(B) If the magnetic field is decreased. Then, according to the formula \(\varepsilon=N \frac{d \phi}{d t}\), the value of the emf is also decreased. The decreases in the emf result in a 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.

According to Lenz law, the direction of the induced current is such a way that to oppose the change in the flux. Thus, the direction of induced current in the triangle is clockwise direction. The direction of the induced current is in the clockwise direction because the decrease in the emf results in a decrease in the flux. Therefore, the induced current will be such that it will increase the flux.


Part A The magnetic flux through the triangle is \(0.000866 \mathrm{~T} \cdot \mathrm{m}^{2}\).

Part B The direction of the induced current is clockwise direction.

Related Solutions

The drawing shows an equilateral triangle, each side of which has a length of 3.99 cm....
The drawing shows an equilateral triangle, each side of which has a length of 3.99 cm. Point charges are fixed to each corner, as shown. The 4.00 C charge experiences a net force due to the charges qA and qB. This net force points vertically downward and has a magnitude of 305 N. Determine (a) charge qA, (b) charge qB.
Find the height of equilateral triangle if side is = 10in
Find the height of equilateral triangle if side is = 10in
A square coil of wire of side 3.00 cm is placed in a uniform magnetic field...
A square coil of wire of side 3.00 cm is placed in a uniform magnetic field of magnitude 1.75 T directed into the page as in the figure shown below. The coil has 38.0 turns and a resistance of 0.780 Ω. If the coil is rotated through an angle of 90.0° about the horizontal axis shown in 0.335 s, find the following. (a) the magnitude of the average emf induced in the coil during this rotation (b) the average current...
Three point charges are located at the corners of an equilateral triangle as in the figure...
Three point charges are located at the corners of an equilateral triangle as in the figure below. Find the magnitude and direction of the net electric force on the 1.70 µC charge. (A = 1.70 µC, B = 7.20 µC, and C = -4.30 µC.) Magnitude N Direction degree (Counterclockwise from the + X-axis)
The loop in the figure below is being pushed into the 0.20 T magnetic field at...
The loop in the figure below is being pushed into the 0.20 T magnetic field at 50 m/s. The resistance of the loop is 0.10 , and its width d = 5.4 cm. What are the direction and the magnitude of the current in the loop? (in units of A) The loop in the figure below is being pushed into the 0.20 T magnetic field at 50 m/s. The resistance of the loop is 0.10 , and its width d...
three charged particles are placed at the corners of an equilateral triangle of side 1.20m. the...
three charged particles are placed at the corners of an equilateral triangle of side 1.20m. the charges are q1 =3.0 q2=-.85uc and q3= -6.0uc. Calculate the magnitude and direction of the net force on each s ide due to the other two. assume the +x axis points to the right that is from q2 to q3. q2 to q3 is along x axis q1 is above it. please show steps.
Three point charges are located on the corners of an equilateral triangle 50cm side: The charge...
Three point charges are located on the corners of an equilateral triangle 50cm side: The charge q_1 = + 10µC is at the origin, The charge q_2 = + 9µC in the upper corner of the triangle, at charge q_3 = -6 µC located at the x axis What is the value of the potential energy of the system of the three charges?
Three point charges are located on the corners of an equilateral triangle 50cm side: The charge...
Three point charges are located on the corners of an equilateral triangle 50cm side: The charge q_1 = + 10µC is at the origin, The charge q_2 = + 9µC in the upper corner of the triangle, at charge q_3 = -6 µC located at the x axis. What is the total force (magnitude and direction) exerted on q_1
An equilateral triangle 7.0 m on a side has a m1 = 25.00 kg mass at...
An equilateral triangle 7.0 m on a side has a m1 = 25.00 kg mass at one corner, a m2 = 85.00 kg mass at another corner, and a m3 = 115.00 kg mass at the third corner. Find the magnitude and direction of the net force acting on the 25.00 kg mass.
three charges are placed on the corners of an equilateral triangle with side lenght L. calculate...
three charges are placed on the corners of an equilateral triangle with side lenght L. calculate the force on the bootom right charge. Record your answer as a vector in i,j,k. your answer should be in term of k,L and Q
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT