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A Long, Straight Wire of Radius R Carries a Current I and is Placed Horizontally in a Uniform Magnetic Field B Pointing Vertically Upward. the Current is Uniformly Distributed Over Its Cross Section.

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प्रश्न

A long, straight wire of radius r carries a current i and is placed horizontally in a uniform magnetic field B pointing vertically upward. The current is uniformly distributed over its cross section. (a) At what points will the resultant magnetic field have maximum magnitude? What will be the maximum magnitude? (b) What will be the minimum  magnitude of the resultant magnetic field?

बेरीज
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उत्तर

(a) As the wire in question is carrying current, so it will also generate a magnetic field around it. And for a long straight wire it will be maximum at the mid-point called P.

Now,

Magnetic field generated by the current carrying wire  \[= \frac{\mu_o i}{2\pi r}\]

Net magnetic field = \[B + \frac{\mu_0 i}{2\pi r}\]


(b) Magnetic field B = 0

when \[r < \frac{\mu_0 i}{2\pi B}\]

Clearly,

B = 0

when

\[r = \frac{\mu_0 i}{2\pi B}\]

But when  \[r > \frac{\mu_0 i}{2\pi B}\]

Net magnetic field = \[B - \frac{\mu_0 i}{2\pi r}\]

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पाठ 35: Magnetic Field due to a Current - Exercises [पृष्ठ २५०]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 35 Magnetic Field due to a Current
Exercises | Q 6 | पृष्ठ २५०

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