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A current carrying toroid winding is internally filled with lithium having susceptibility χ = 2.1 × 10−5. What is the percentage increase in the magnetic field in the presence of lithium over that - Physics

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प्रश्न

A current carrying toroid winding is internally filled with lithium having susceptibility `chi` = 2.1 × 10−5. What is the percentage increase in the magnetic field in the presence of lithium over that without it?
संख्यात्मक

उत्तर

Given:

`chi` = 2.1 × 10−5

To find:

% increase in the magnetic field = ?

We know the magnetic field inside the toroid without lithium

B0 = μ0nI

B0 = μ0H   ...(i)

The magnetic field inside the toroid with lithium

B = µH    ...(ii)

∴ B − B0 = μH − μ0H

= (μ − μ0)H

∴ The percentage increase in the magnetic field after inserting lithiu m is

% increase in magnetic field

= `(B - B_0)/B_0 xx 100`

= `((mu - mu_0)H)/(mu_0H) xx 100`

% increase in magnetic field

= `(mu - mu_0)/mu_0 xx 100`

But `(mu - mu_0)/mu_0 = chi`

∴ % increase in magnetic field

= `chi xx 100 = 2.1 xx 10^-5 xx 100`

= 0.0021%

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Magnetic Field of a Solenoid and a Toroid
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