Solveeit Logo
Login

Question

Question: Two circular coils of radii \(5\,cm\) and \(10\,cm\) carrying equal currents of \(2\,A\).The coils h...

Two circular coils of radii 5 cm5\,cm and 10 cm10\,cm carrying equal currents of 2 A2\,A.The coils have 5050 and 100100 turns respectively and are placed in such a way that their planes as well as their centers coincide. Magnitude of magnetic field at the common center coil is:
A. 8π×10−6T8\pi \times {10^{ - 6}}T if current in the coils are in the same sense.
B. 4π×10−6T4\pi \times {10^{ - 6}}T if the current in the coils is in the opposite sense.
C. Zero if current in the coils are in the opposite sense.
D. 8π×10−6T8\pi \times {10^{ - 6}}T if current in the coils are in the opposite sense.

Explanation

Solution

From the given data, we have to find out the magnitude of magnetic field at the common center coil, for this calculation they have given radius and current and number of turns of a coil. For this calculation we are using the magnetic field equation.

Complete step by step answer:
A straight current-carrying conductor has a magnetic field in the shape of concentric circles around it. The magnetic field of a straight current-carrying conductor can be visualized by magnetic field lines. The direction of a magnetic field produced due to a current-carrying conductor relies upon the same direction in which the current is flowing.Here,
r1=5 cm ⇒r2=10 cm ⇒n1=50 ⇒n2=100 ⇒i=2 A {r_1} = 5\,cm \\\ \Rightarrow {r_2} = 10\,cm \\\ \Rightarrow {n_1} = 50 \\\ \Rightarrow {n_2} = 100 \\\ \Rightarrow i = 2\,A \\\
When the current coils in the same direction, then from magnetic field,
B=B1+B2B = {B_1} + {B_2}
magnetic field formula,
{B_1} = \dfrac{{{\mu _ \circ }i{n_1}}}{{2{r_1}}} \\\ \Rightarrow {B_2} = \dfrac{{{\mu _ \circ }i{n_2}}}{{2{r_2}}} \\\
Substituting the B1{B_1}and B2{B_2}in the magnetic field, then
B=μ∘in12r1+μ∘in22r2→(1)B = \dfrac{{{\mu _ \circ }i{n_1}}}{{2{r_1}}} + \dfrac{{{\mu _ \circ }i{n_2}}}{{2{r_2}}} \to (1)
Here we know μ∘=4π×10−7{\mu _ \circ } = 4\pi \times {10^{ - 7}}
By adding all the value in equation (1),
B=4π×10−7×50×22×0.05+4π×10−7×100×22×0.1B = \dfrac{{4\pi \times {{10}^{ - 7}} \times 50 \times 2}}{{2 \times 0.05}} + \dfrac{{4\pi \times {{10}^{ - 7}} \times 100 \times 2}}{{2 \times 0.1}}
Calculating the above values we get the value of B,
B=8π×10−6B = 8\pi \times {10^{ - 6}}
Now we clear that, current in the coils is in the same sense.

Lets calculate the magnetic field in opposite direction
B=B1−B2B = {B_1} - {B_2}
By adding the same values in opposite direction, we get the value of B
B=4π×10−7×50×22×0.05−4π×10−7×100×22×0.1B = \dfrac{{4\pi \times {{10}^{ - 7}} \times 50 \times 2}}{{2 \times 0.05}} - \dfrac{{4\pi \times {{10}^{ - 7}} \times 100 \times 2}}{{2 \times 0.1}}
Calculating the above value , B=0B = 0. We have proven that the current in the coils are in the opposite sense. Magnitude of the magnetic field at the common center point is proven in the same sense and opposite sense.

Hence, both options A and C are correct.

Note: Circular coils are in the same sense or in opposite sense we have to prove in two forms. Magnitude of magnetic field ranges from approximately 25 to 65 microtales. Direction of magnetic field defined by right hand rule.magnetic field is which repels the magnet to each other.