Maharashtra State BoardHSC Science (Electronics) 11th
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Solve the following problem. A magnetic pole of bar magnet with pole strength of 100 A m is 20 cm away from the centre of a bar magnet. - Physics

Sum

Solve the following problem.

A magnetic pole of a bar magnet with a pole strength of 100 A m is 20 cm away from the centre of a bar magnet. The bar magnet has a pole strength of 200 A m and has a length of 5 cm. If the magnetic pole is on the axis of the bar magnet, find the force on the magnetic pole.

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Solution

Given that, (qm)1 = 200 Am and
(2l) = 5 cm = 5 × 10–2 m
∴ m = 200 × 5 × 10–2 = 10 Am2
For a bar magnet, magnetic dipole moment is,
m = qm (2l)
For a point on the axis of a bar magnet at distance, r = 20 cm = 0.2 m,
Ba = `μ_0/(4π)xx(2"m")/"r"^3`

= `10^-7xx(2xx10)/(0.2)^3`

= 0.25 × 10−3
= 2.5 × 10−4 Wb/m2
The force acting on the pole will be given by,
F = qm Ba = 100 × 2.5 × 10–4
= 2.5 × 10–2 N

The force acting on the magnetic pole due to the bar magnet is 2.5 × 10–2 N.

Concept: The Bar Magnet
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APPEARS IN

Balbharati Physics 11th Standard Maharashtra State Board
Chapter 12 Magnetism
Exercises | Q 4. (i) | Page 228
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