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A long solenoid has 200 turns per cm and carriers a currenti. The magnetic field at its centre is $6.28 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^2$. Another long solenoid has 100 turns per $\mathrm{cm}$ and it carries a current $\frac{i}{3}$. The value of magnetic field at its centre is nearly
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The correct answer is:
$1.05 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^2$
Magnetic field due to a long solenoid is given by
$B=\mu_0 n i$
For solenoid with 200 turns per $\mathrm{cm}$ and $i$ current flows through it,
$6.28 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^2=\mu_0 \times 200 \times i$
Now, for solenoid with 100 turns per cm with current $\frac{i}{3}$,
$B=\mu_0 \times 100 \times\left(\frac{i}{3}\right)$
Introducing value of $i$ from equation (1),
$B=\mu_0 \times 100 \times\left(\frac{1}{3} \times \frac{6.28 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^2}{\mu_0 \times 200}\right)=1.46 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^2$
$B \approx 1.05 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^2$
$B=\mu_0 n i$
For solenoid with 200 turns per $\mathrm{cm}$ and $i$ current flows through it,
$6.28 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^2=\mu_0 \times 200 \times i$
Now, for solenoid with 100 turns per cm with current $\frac{i}{3}$,
$B=\mu_0 \times 100 \times\left(\frac{i}{3}\right)$
Introducing value of $i$ from equation (1),
$B=\mu_0 \times 100 \times\left(\frac{1}{3} \times \frac{6.28 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^2}{\mu_0 \times 200}\right)=1.46 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^2$
$B \approx 1.05 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^2$
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