question_answer

A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is:                                                                                                            [NEET 2020] \[\left( {{\mu }_{0}}=4\pi \times {{10}^{-7}}T\,m\,{{A}^{-1}} \right)\]

A)  \[3.14\times {{10}^{-4}}T\]     

B)       \[6.28\times {{10}^{-5}}T\]

C)  \[3.14\times {{10}^{-5}}T\]     

D)       \[6.28\times {{10}^{-4}}T\]

Correct Answer: D Solution : [d] Magnetic field at centre of solenoid \[{{\mu }_{0}}nl\]

\[n=\frac{N}{L}=\frac{100}{50\times {{10}^{-2}}}=200\,\text{turns/m}\]

\[\text{l = 2}\text{.5 A}\]

On putting the values

\[\text{B = 4}\pi \times \text{1}{{\text{0}}^{-7}}\times 200\times 2.5\]

\[=6.28\times {{10}^{-7}}T\]