A wire carrying current $I$ has the shape as shown in adjoining figure.Linear parts of the wire are very long and parallel to $X-$axis while semicircular portion of radius $R$ is lying in $Y-Z$ plane. Magnetic field at point $O$ is
A wire carrying current $I$ has the shape as shown in adjoining figure.Linear parts of the wire are very long and parallel to $X-$axis while semicircular portion of radius $R$ is lying in $Y-Z$ plane. Magnetic field at point $O$ is
- [AIPMT 2015]
- A
$\overrightarrow {\;B} = \frac{{{\mu _0}I}}{{4\pi R}}\left( {\pi \hat i + 2\hat k} \right)\;\;\;\;\;$
- B
$\;\overrightarrow {\;B} = - \frac{{{\mu _0}I}}{{4\pi R}}\left( {\pi \hat i - 2\hat k} \right)$
- C
$\;\overrightarrow {\;B} = - \frac{{{\mu _0}I}}{{4\pi R}}\left( {\pi \hat i + 2\hat k} \right)$
- D
$\;\overrightarrow {\;B} = \frac{{{\mu _0}I}}{{4\pi R}}\left( {\pi \hat i - 2\hat k} \right)$
Similar Questions
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- [IIT 2012]
An infinitely long conductor $PQR$ is bent to form a right angle as shown. A current $I$ flows through $PQR$ The magnetic field due to this current at the point $M $ is $H_1$. Now another infinitely long straight conductor $QS$ is connected at $Q$ so that the current is $I/2$ in $QR$ as well as in $QS$, The current in $PQ$ remaining unchanged. The magnetic field at $M$ is now ${H_{2.}}$The ratio ${H_1}/{H_2}$ is given by
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- [IIT 2000]
Two circular coils $P$ and $Q$of $100$ turns each have same radius of $\pi \mathrm{cm}$. The currents in $\mathrm{P}$ and $\mathrm{R}$ are $1 \mathrm{~A}$ and $2 \mathrm{~A}$ respectively. $\mathrm{P}$ and $\mathrm{Q}$ are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is $\sqrt{\mathrm{x}} \mathrm{mT}$, where X=___.
$\left[\text { Use } \mu_0=4 \pi \times 10^{-7} \mathrm{TmA}^{-1}\right]$
Two circular coils $P$ and $Q$of $100$ turns each have same radius of $\pi \mathrm{cm}$. The currents in $\mathrm{P}$ and $\mathrm{R}$ are $1 \mathrm{~A}$ and $2 \mathrm{~A}$ respectively. $\mathrm{P}$ and $\mathrm{Q}$ are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is $\sqrt{\mathrm{x}} \mathrm{mT}$, where X=___.
$\left[\text { Use } \mu_0=4 \pi \times 10^{-7} \mathrm{TmA}^{-1}\right]$
- [JEE MAIN 2024]