Q183 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Physics
If the current 'i' is passing through the wire, it has a shape as shown in the figure. The radius of the curved part is R and the linear part is assumed to be very long. The magnetic induction of the field at the point O, is
($\mu_0=$permeability of free space)
($\mu_0=$permeability of free space)
- A$\dfrac{\mu_0 i}{2\pi R}\left(2+\dfrac{\pi}{2}\right)$
- Correct answerB$\dfrac{\mu_0 i}{4\pi R}\left(2+\pi\right)$
- C$\dfrac{\mu_0 i}{2\pi R}\left(1+\dfrac{3\pi}{2}\right)$
- D$\dfrac{\mu_0 i}{\pi R}\left(1+\dfrac{\pi}{2}\right)$
Solution
The wire can be divided into three parts: a lower semi-infinite straight wire, a semicircular arc, and an upper semi-infinite straight wire.
The magnetic field at point O due to the lower semi-infinite straight wire is given by $B_1 = \dfrac{\mu_0 i}{4\pi R}$. Applying the right-hand rule, the current is towards the left and point O is above the wire, so the magnetic field is directed into the page.
The magnetic field at point O due to the semicircular arc is given by $B_2 = \dfrac{\mu_0 i}{4R} = \dfrac{\pi \mu_0 i}{4\pi R}$. The current is in the clockwise direction, so the magnetic field is directed into the page.
The magnetic field at point O due to the upper semi-infinite straight wire is given by $B_3 = \dfrac{\mu_0 i}{4\pi R}$. The current is towards the right and point O is below the wire, so the magnetic field is directed into the page.
Since all three magnetic field components are in the same direction, the total magnetic field at O is their sum:
$B_{net} = B_1 + B_2 + B_3$
$B_{net} = \dfrac{\mu_0 i}{4\pi R} + \dfrac{\pi \mu_0 i}{4\pi R} + \dfrac{\mu_0 i}{4\pi R}$
$B_{net} = \dfrac{\mu_0 i}{4\pi R}(1 + \pi + 1)$
$B_{net} = \dfrac{\mu_0 i}{4\pi R}(2 + \pi)$
The magnetic field at point O due to the lower semi-infinite straight wire is given by $B_1 = \dfrac{\mu_0 i}{4\pi R}$. Applying the right-hand rule, the current is towards the left and point O is above the wire, so the magnetic field is directed into the page.
The magnetic field at point O due to the semicircular arc is given by $B_2 = \dfrac{\mu_0 i}{4R} = \dfrac{\pi \mu_0 i}{4\pi R}$. The current is in the clockwise direction, so the magnetic field is directed into the page.
The magnetic field at point O due to the upper semi-infinite straight wire is given by $B_3 = \dfrac{\mu_0 i}{4\pi R}$. The current is towards the right and point O is below the wire, so the magnetic field is directed into the page.
Since all three magnetic field components are in the same direction, the total magnetic field at O is their sum:
$B_{net} = B_1 + B_2 + B_3$
$B_{net} = \dfrac{\mu_0 i}{4\pi R} + \dfrac{\pi \mu_0 i}{4\pi R} + \dfrac{\mu_0 i}{4\pi R}$
$B_{net} = \dfrac{\mu_0 i}{4\pi R}(1 + \pi + 1)$
$B_{net} = \dfrac{\mu_0 i}{4\pi R}(2 + \pi)$
Hence, the correct answer is option: 'B'.