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The Coercive Force for a Certain Permanent Magnet is 4.0 × 104 a M−1. this Magnet is Placed Inside a Long Solenoid of 40 Turns/Cm and a Current is Passed in the Solenoid to Demagnetise It Completely. - Physics

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प्रश्न

The coercive force for a certain permanent magnet is 4.0 × 104 A m−1. This magnet is placed inside a long solenoid of 40 turns/cm and a current is passed in the solenoid to demagnetise it completely. Find the current.

योग
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उत्तर

Given:-

Number of turns per unit length, n = 40 turns/cm = 4000 turns/m

Magnetising field, H = 4 × 104 A/m

Magnetic field inside a solenoid (B) is given by,

B = µ0nI,

where, n = number of turns per unit length.

I = current through the solenoid.

\[\therefore   \frac{B}{\mu_0}   =   nI   =   H\]

\[\Rightarrow H = \frac{N}{l}I\]

\[ \Rightarrow I = \frac{Hl}{N}   =   \frac{H}{n}\]

\[ \Rightarrow I = \frac{4 \times {10}^4}{4000} = 10  A\]

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अध्याय 15: Magnetic Properties of Matter - Exercises [पृष्ठ २८७]

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एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
अध्याय 15 Magnetic Properties of Matter
Exercises | Q 9 | पृष्ठ २८७

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