How many times will the magnetic field induction in the center of a ring with current decrease if it is bent at an angle \alpha? The current in the ring does not change.
Solutions of Savchenko Problems in Physics knowledge must be free
<h3 id="back-link"><a href="/#9.2">$\leftarrow$Back</a></h3>
<h3> Statement </h3>
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$9.2.11.$ How many times will the magnetic field induction in the center of a ring with current decrease if it is bent at an angle $\alpha$? The current in the ring does not change.
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<h3>Solution</h3>
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Magnetic field induction at the center of a bent ring
In 9.2.10 we obtained an expression for the magnetic induction at the center of the ring
$$B_0 = \frac{\mu_0 I}{2R}\quad(1)$$
Due to superposition, the magnetic induction at the center of one semicircle will be half of this value:
$$B_1 = B_2 = \frac{B_0}{2} = \frac{\mu_0 I}{4R}\quad(2)$$
From the drawing, it is clear that the angle between the vectors $\vec{B}_1$ and $\vec{B}_2$ is equal to $\alpha$$$(\vec{B_1},\vec{B_2})=\alpha$$
The total vector of magnetic induction is equal to the vector sum of the components of magnetic induction of two rings $\vec{B}_1$ and $\vec{B}_2$$$\vec{B} = \vec{B_1} + \vec{B_2}$$
We will find the modulus of this vector using the cosine law.
$$B = \sqrt{{B_1}^2 + {B_2}^2 - 2B_1 B_2 \cos\alpha}$$
Taking into account the expression $(2)$, we find the total magnetic induction as
$$B= \frac{\mu_0 I}{4R}\sqrt{2-2\cos\alpha} = \frac{\mu_0 I}{2R}\cdot\sin\frac{\alpha}{2}\quad(3)$$
Dividing the expression $(3)$ by $(1)$, we obtain the desired ratio
$$\boxed{\frac{B}{B_0} = \sin\frac{\alpha}{2}}$$
Savchenko Solutions
