Q24E - University Physics with Modern Physics

Question

A rectangular loop with dimensions 4.20 cm by 9.50 cm carries current . The current in the loop produces a magnetic field at the center of the loop that has magnitude 5.50$\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathit{T}$ and direction away from you as you view the plane of the loop. What are the magnitude and direction (clockwise or counter clockwise) of the current in the loop?

Step-by-Step Solution

Answer

The magnitude of the current in the loop is 2.64 A . The direction of the current in the loop is clockwise.

Step 1: The magnetic field due to one side of the rectangular loop at center

The magnetic field due to one side of the rectangular loop at the center is given by: 

 

 $B=\frac{{\mu }_{0}l}{4\pi }\frac{2l}{x\sqrt{x^2+l^2}}$

 

Where, l is the half-length of the wire, x is the distance from the wire from the center of the loop.

 

So,

 

The magnetic field due to the short side a of the rectangular loop at the center is given by: 

$B_a=\frac{{\mu }_{0}l}{4\pi }\frac{2a}{b\sqrt{\frac{b^2}{4}+\frac{a^2}{4}}}$ 

 

Here, x is equal to  $\frac{a}{2}$ and l is equal to $\frac{b}{2}$ .

 

The magnetic field due to the longer side b of the rectangular loop at the center is given by :

 

$B_b=\frac{{\mu }_{0}I}{4\pi }\frac{2b}{a\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}}$ 

Here, x is equal to  $\frac{b}{2}$ and l is equal to $\frac{a}{2}$ .

 

The total magnetic field at the center is due to four wires at the center of the rectangular loop: 

 

The magnetic field at the center due to two wires of length a and two wires of length b is given by the vector sum of the magnetic field of four wires at that center.

 

So, it is given by: 

 

 $B=2B_a+2B_b$

 

The direction of current flow in wire and originating magnetic field:

 

The direction of the magnetic field due to the current-carrying conductor can be given by the right-hand thumb rule,

According to the right-hand thumb rule, if the thumb of the right-hand points along the direction of the current, then the remaining curled fingers of the same hand give the direction of the magnetic field due to the current.

Step 2: The direction of current flow in a rectangular loop

Given data: 

 

• The shorter side of the loop is 4.20 cm. 
• The longer side of the loop is 9.50 cm. 

The magnetic field at the center of the loop is 5.50×10^-5 T

Here, the direction of the magnetic field due to the current-carrying rectangular loop is inward of the plane; therefore, according to the right-hand thumb rule current must be flow in the clockwise direction in the loop. 

 

Thus, the direction of the current in the loop is clockwise.

Step 3: Calculation of current in a rectangular loop

Using formula:

 

$\begin{array}{r}B=2B_a+2B_b\\ B=2\left(B_a+B_b\right)\end{array}$

 

Now, putting the values and we get,

 $\begin{array}{r}B=2\left(\frac{{\mu }_{0}I}{4\pi }\frac{2a}{b\sqrt{\frac{b^2}{4}+\frac{a^2}{4}}}+\frac{{\mu }_{0}I}{4\pi }\frac{2b}{a\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}}\right)\\ B=\left(\frac{2{\mu }_{0}I}{\pi }\right)\left(\frac{a}{b\sqrt{b^2+a^2}}+\frac{b}{a\sqrt{b^2+a^2}}\right) \end{array}$

 Now, putting the values of constants in the above equation, and we get,

 $\begin{array}{r}5.50\times {10}^{-5}\mathrm{T}=\left(\frac{2\times \left(4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)\times \mathrm{I}}{\pi }\right)\left(\begin{array}{c}\frac{0.0420\mathrm{m}}{\left(0.0950\mathrm{m}\right)\sqrt{(0.0950\mathrm{m}{)}^2+(0.0420\mathrm{m}{)}^2}}\\ \frac{0.0950\mathrm{m}}{0.0420\mathrm{m}\sqrt{(0.0950\mathrm{m}{)}^2+(0.0420\mathrm{m}{)}^2}}\end{array}\right)\\ I=\frac{5.50\times {10}^{-5}\mathrm{T}}{2.082\times {10}^{-5}\mathrm{T}/\mathrm{A}} \\ I=2.64\mathrm{A} \end{array}$

Thus, the magnitude of the current in the loop is .