Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

41 of 276

Question 41 of 276

The magnetic intensity at the centre of a long current carrying solenoid is found to be 1.6×103Am−11.6\times {10}^{3}A{m}^{-1}. If the number of turns is 88 per cmcm, then the current flowing through the solenoid is ______A.

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Answer: 2

The magnetic intensity HH is defined as the product of the number of turns per unit length in a coil nn and the current that it carriesII.
The magnetic field of a solenoid is given by B=μ0niB={\mu }_{0}ni
H=Bμ0=μ0niμ0⇒H=ni...(i)H=\dfrac{B}{{\mu }_{0}}=\dfrac{{\mu }_{0}ni}{{\mu }_{0}} \Rightarrow H=ni...(i)
The given data is
H=1.6×103Am−1H=1.6\times {10}^{3}A{m}^{-1}
n=8n=8 per cmcm
Substituting the values in equation (i)
⇒1.6×1000=81100×i\Rightarrow 1.6\times 1000=\dfrac{8}{\dfrac{1}{100}}\times i
⇒i=2A\Rightarrow i=2A

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