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A toroidal solenoid has 3000 turns and a mean radius of 10 cm. it has a soft iron core of relative permeability 2000 . What is the magnitude of the magnetic field in the core when a current of 1 A is passed through the solenoid?
 

Option: 1

0.012 \mathrm{~T}


Option: 2

0.12 \mathrm{~T}


Option: 3

1.2 \mathrm{~T}


Option: 4

12 \mathrm{~T}


Answers (1)

best_answer

The magnetic field in the core is given by

\mathrm{B=\mu n I}

where \mathrm{\mu}  is the permeability of soft iron and n is the number of turns per unit length of the solenoid. Now

\mathrm{\mu_r =\frac{\mu}{\mu_0} \text { and } n=\frac{3000}{2 \pi r}=\frac{3000}{2 \pi \times 0.1} }
\mathrm{\therefore \quad B =\mu_r \mu_0 n I \\ }
\mathrm{ =2000 \times 4 \pi \times 10^{-7} \times \frac{3000}{2 \pi \times 0.1} \times 1=12 \mathrm{~T}}

Hence the correct choice is (d).

Posted by

Gaurav

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