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प्रश्न
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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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संबंधित प्रश्न
A solenoid of length π m and 5 cm in diameter has winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its center along the axis.
A toroid of a central radius of 10 cm has windings of 1000 turns. For a magnetic field of 5 × 10-2 T along its central axis, what current is required to be passed through its windings?
What is Solenoid?
What is Toroid?
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