# Question Video: Calculating the Magnetic Dipole Moment of a Circular Coil Physics

A circular coil of diameter 12 cm carries a current πΌ that generates a magnetic field in its center of magnitude 2 Γ 10β»β΄ T. Given that πβ = 4π Γ 10β»β· H/m, calculate the magnetic dipole moment of this coil.

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### Video Transcript

A circular coil of diameter 12 centimeters carries a current πΌ that generates a magnetic field in its center of magnitude two times 10 to the power of negative four teslas. Given that π naught equals four π times 10 to the power of negative seven henries per meter, calculate the magnetic dipole moment of this coil. (A) 1.782 ampere meters squared, (B) 3.6 ampere meters squared, (C) 0.0624 ampere meters squared, (D) 0.216 ampere meters squared.

In this question, we are being asked to calculate the magnetic dipole moment of this coil. We can recall that if we have a loop of current-carrying wire, where the current has a magnitude πΌ and the area of that loop is called π΄, then the magnetic dipole moment of the loop π sub π is equal to πΌ times π΄. We are not given a value for the current πΌ, so weβll have to use an equation substitution that uses values we are given in the question.

We are told in the question that the circular coil generates a magnetic field in its center. Now, we can recall that the magnetic field strength π΅ at the center of a current-carrying loop of wire is given by the equation π΅ equals π naught πΌ over two π, where πΌ is the current in the loop, π is the radius of the loop, and π naught is the permeability of free space.

Now, we can rearrange this equation to make the current πΌ the subject. We can multiply both sides of the equation by two π, and then divide both sides by π naught, to leave us with the equation πΌ equals two ππ΅ over π naught. Substituting this into our equation for π sub π, we have π sub π equals two ππ΅ over π naught multiplied by π΄.

We can now find an expression for the area π΄. The area of the circular coil is the area of a circle, which is equal to ππ squared, where π is the radius of the circle. Substituting this into our equation for π sub π, we have π sub π equals two ππ΅ over π naught multiplied by ππ squared. Alright then, we have our final equation. So, letβs write out what values we have for each variable.

The ones we are explicitly given are the magnetic field strength, π΅, as two times 10 to the power of negative four teslas and the permeability of free space, π naught, as four π times 10 to the power of negative seven henries per meter. Then, we are given the diameter of the loop of wire, 12 centimeters, but what we need is its radius, π. Before going further, letβs put this value of diameter into the SI units of meters. We do this by dividing 12 centimeters by 100, as there are 100 centimeters in one meter. This gives us a value of 0.12 meters for the diameter.

Now then, the radius of the loop is just half of its diameter. So, the radius is equal to 0.06 meters. Finally, with all of our variables known, we can substitute them into the equation for the magnetic dipole moment as follows. The units of teslas multiplied by meters squared over henries gives us units of amperes. So, our final answer will be in ampere meters squared. Solving the equation in our calculators then, we get a value of 0.216 ampere meters squared for the magnetic dipole moment of this coil.

If we now look at our answer options, we can see that this answer corresponds with option (D). So, we can eliminate options (A), (B), and (C) and confirm that option (D), 0.216 ampere meters squared, is the correct answer for the magnetic dipole moment of this coil.