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The magnetic field inside a current carrying toroidal solenoid is $0.2 \mathrm{mT}$. What is the magnetic field inside the toroid if the current
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Verified Answer
The correct answer is:
$0.6 \mathrm{mT}$
Magnetic field inside the toroid is given by
$$
\begin{aligned}
& \mathrm{B}=\mu_0 \mathrm{nI} \\
& \therefore \frac{\mathrm{B}_2}{\mathrm{~B}_1}=\frac{\mathrm{I}_2}{\mathrm{I}_1}=3
\end{aligned}
$$
(If cross-sectional radius of the solenoid is changed three will be no effect on the magnetic field, assuming the number of turns remains same.)
$$
\therefore \mathrm{B}_2=3 \mathrm{~B}_1=3 \times 0.2=0.6 \mathrm{~T}
$$
$$
\begin{aligned}
& \mathrm{B}=\mu_0 \mathrm{nI} \\
& \therefore \frac{\mathrm{B}_2}{\mathrm{~B}_1}=\frac{\mathrm{I}_2}{\mathrm{I}_1}=3
\end{aligned}
$$
(If cross-sectional radius of the solenoid is changed three will be no effect on the magnetic field, assuming the number of turns remains same.)
$$
\therefore \mathrm{B}_2=3 \mathrm{~B}_1=3 \times 0.2=0.6 \mathrm{~T}
$$
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