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For a wire, as shown in the figure, carrying a current of $10 \mathrm{~A}$, then the magnetic induction field at the point $O$ is [talse, $\mu_0=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}$ ]
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Verified Answer
The correct answer is:
$4 \times 10^4 \mathrm{~T}$
Magnetic field due to a Iong current carrying wire at one end is given by $B=\frac{\mu_0 i}{4 \pi d}$.
(where, $i=$ current and $d=$ distance from the wire) For both the wire,
$$
i=10 \mathrm{~A}, d=\frac{1}{2} \times 10^{-2} \mathrm{~m}
$$
Magnetic field due to one wire,
$$
\begin{gathered}
B=\frac{\mu_0 \times 10}{4 \pi \times\left(\frac{1}{2} \times 10^{-2}\right)} \\
B=\frac{2 \times 10^{-7} \times 10}{10^{-2}}=2 \times 10^{-4} \mathrm{~T}
\end{gathered}
$$
Due to both wires magnetic field is in same direction, so they will add up.
$\Rightarrow$ Total magnetic field, $B=2\left(2 \times 10^{-4}\right)$
$$
B=4 \times 10^{-4} \mathrm{~T}
$$
(where, $i=$ current and $d=$ distance from the wire) For both the wire,
$$
i=10 \mathrm{~A}, d=\frac{1}{2} \times 10^{-2} \mathrm{~m}
$$
Magnetic field due to one wire,
$$
\begin{gathered}
B=\frac{\mu_0 \times 10}{4 \pi \times\left(\frac{1}{2} \times 10^{-2}\right)} \\
B=\frac{2 \times 10^{-7} \times 10}{10^{-2}}=2 \times 10^{-4} \mathrm{~T}
\end{gathered}
$$
Due to both wires magnetic field is in same direction, so they will add up.
$\Rightarrow$ Total magnetic field, $B=2\left(2 \times 10^{-4}\right)$
$$
B=4 \times 10^{-4} \mathrm{~T}
$$
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