ACROSS \[AB\] \[{{i}_{1}}{{R}_{1}}={{i}_{2}}{{R}_{2}}\Rightarrow {{i}_{1}}{{l}_{2}}={{i}_{2}}{{l}_{2}}\] \[\left( \because \ R=\rho \frac{l}{A} \right)\] Also \[{{B}_{1}}=\frac{{{\mu }_{o}}}{4\pi }\times \frac{{{i}_{1}}{{l}_{1}}}{{{r}^{2}}}\] and \[{{B}_{2}}=\frac{{{\mu }_{o}}}{4\pi }\times \frac{{{i}_{2}}{{l}_{2}}}{{{r}^{2}}}\] (\[\because \ l=r\theta \])  \[\therefore \,\frac{{{B}_{2}}}{{{B}_{1}}}=\frac{{{i}_{1}}{{l}_{1}}}{{{i}_{2}}{{l}_{2}}}=1\] Hence, two FIELD induction?s are equal but of opposite direction. So, resultant magnetic induction at the centre is zero and is independent of\[\theta \].

"> ACROSS \[AB\] \[{{i}_{1}}{{R}_{1}}={{i}_{2}}{{R}_{2}}\Rightarrow {{i}_{1}}{{l}_{2}}={{i}_{2}}{{l}_{2}}\] \[\left( \because \ R=\rho \frac{l}{A} \right)\] Also \[{{B}_{1}}=\frac{{{\mu }_{o}}}{4\pi }\times \frac{{{i}_{1}}{{l}_{1}}}{{{r}^{2}}}\] and \[{{B}_{2}}=\frac{{{\mu }_{o}}}{4\pi }\times \frac{{{i}_{2}}{{l}_{2}}}{{{r}^{2}}}\] (\[\because \ l=r\theta \])  \[\therefore \,\frac{{{B}_{2}}}{{{B}_{1}}}=\frac{{{i}_{1}}{{l}_{1}}}{{{i}_{2}}{{l}_{2}}}=1\] Hence, two FIELD induction?s are equal but of opposite direction. So, resultant magnetic induction at the centre is zero and is independent of\[\theta \].

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A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle \[\theta \] at the centre. The value of the magnetic induction at the centre due to the current in the ring is [IIT 1995]

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) 1 year ago

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Directions of currents in two parts are different, so directions of magnetic fields due to these currents are opposite. Also applying Ohm?s law ACROSS \[AB\] \[{{i}_{1}}{{R}_{1}}={{i}_{2}}{{R}_{2}}\Rightarrow {{i}_{1}}{{l}_{2}}={{i}_{2}}{{l}_{2}}\] \[\left( \because \ R=\rho \frac{l}{A} \right)\] Also \[{{B}_{1}}=\frac{{{\mu }_{o}}}{4\pi }\times \frac{{{i}_{1}}{{l}_{1}}}{{{r}^{2}}}\] and \[{{B}_{2}}=\frac{{{\mu }_{o}}}{4\pi }\times \frac{{{i}_{2}}{{l}_{2}}}{{{r}^{2}}}\] (\[\because \ l=r\theta \])  \[\therefore \,\frac{{{B}_{2}}}{{{B}_{1}}}=\frac{{{i}_{1}}{{l}_{1}}}{{{i}_{2}}{{l}_{2}}}=1\] Hence, two FIELD induction?s are equal but of opposite direction. So, resultant magnetic induction at the centre is zero and is independent of\[\theta \].

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Posted on 12 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Physics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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