question_answer

Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5 A. (See figure) [JEE Main 12-4-2019 Afternoon]

A) \[3.0\times {{10}^{5}}T\]   

B) \[2.5\times {{10}^{5}}T\]

C) \[2.0\times {{10}^{5}}T\]   

D)   \[1.5\times {{10}^{5}}T\]

Correct Answer: D

Solution :

          \[B=\frac{{{\mu }_{0}}I}{4\pi d}2\sin \theta \] \[d=4cm\] \[\sin \theta =\frac{3}{5}\]