Question

The magnetic moment of a thin round loop with current, if the radius of the loop is equal to $R = 100mm$ and the magnetic field at its center is equal to $B = 6.0\mu T$, is $30 \times {10^{ - x}}A - {m^2}$. Find the value of x.

Answer 1

Hint: First, we will first find out the value of i through the formula of magnetic field. Then we will put that value of current into the formula of magnetic moment and then solve it further by expanding the formula of area and substituting it. Then we will get a numerical value of the magnetic moment and compare it with the one given in the question. Refer to the solution below.

Complete step-by-step answer:
Formula used: $M = iA$, $B = \dfrac{{\mu i}}{{2r}}$

As we know that the formula for magnetic moment is-

$\Rightarrow M = iA$
Where, i stands for the value of current in wire multiplied by A, area of the loop.
The value of current is not given but the value of magnetic field and the radius of the loop is given.

The formula of the magnetic field due to a circular loop is $B = \dfrac{{\mu i}}{{2r}}$.

$\Rightarrow B = \dfrac{{{\mu _0}i}}{{2r}} \\\ \\\ \Rightarrow i = \dfrac{{B2r}}{{{\mu _0}}} \\\$

Putting the value of i in the formula for magnetic moment mentioned above, we get-

$\Rightarrow M = iA \\\ \\\ \Rightarrow M = \dfrac{{B2r}}{{{\mu _0}}}A \\\$

The formula for area is $A = \pi {r^2}$

$\Rightarrow M = \dfrac{{B2r}}{{{\mu _0}}}A \\\ \\\ \Rightarrow M = \dfrac{{B2r}}{{{\mu _0}}}.\pi {R^2} \\\$

Now, multiplying the numerator and the denominator by 2, in order to make the expression as $\dfrac{{4\pi }}{{{\mu _0}}}$. We do this because we know that the value of $\dfrac{{{\mu _0}}}{{4\pi }}$ is ${10^{ - 7}}$. Thus, the value of $\dfrac{{4\pi }}{{{\mu _0}}}$ will be $\dfrac{1}{{{{10}^{ - 7}}}}$. Hence-

$\Rightarrow M = \dfrac{{B2r}}{{{\mu _0}}}.\pi {R^2} \\\ \\\ \Rightarrow M = \dfrac{{4\pi \times B{R^3}}}{{{\mu _0} \times 2}} \\\ \\\ \Rightarrow M = \dfrac{{6 \times {{10}^{ - 6}} \times {{\left( {{{10}^{ - 1}}} \right)}^3}}}{{{{10}^{ - 7}} \times 2}} \\\ \\\ \Rightarrow M = \dfrac{{3 \times {{10}^{ - 6}} \times {{\left( {{{10}^{ - 1}}} \right)}^3}}}{{{{10}^{ - 7}}}} \\\ \\\ \Rightarrow M = \dfrac{{3 \times {{10}^{ - 6}} \times {{10}^{ - 3}}}}{{{{10}^{ - 7}}}} \\\ \\\ \Rightarrow M = 3 \times {10^{ - 6}} \times {10^{ - 3}} \times {10^7} \\\ \\\ \Rightarrow M = 30 \times {10^{ - 9}} \times {10^6} \\\ \\\ \Rightarrow M = 30 \times {10^{ - 3}}A - {m^2} \\\$

Now, the value of magnetic moment as per given in the question was $30 \times {10^{ - x}}A - {m^2}$. Comparing both the values, we get-

$\Rightarrow 30 \times {10^{ - x}}A - {m^2} =$ $30 \times {10^{ - 3}}A - {m^2}$

Hence, the value of x is 3.

Note: The magnet moment reflects the electrical strength and direction of an electrical or some other entity generating a magnet effect. Electrical current loops, including electromagnets, permanent magnets, traveling elementary particles (such as electrons) and various molecules and other astronomical bodies (such as asteroids, other planets, stars, etc.) are examples of structures of magnetic moments.