Magnetic properties of matter

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Question

The percentage increase in the magnetic field B when the space within a current carrying toroid is filled with aluminium, with a susceptibility of $2.1 \times 10^{- 5}$ , is $2.1 \times 10^{- p} .$ Find $p$ .

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Solution

In the absence of aluminium the magnetic field is $B_{0} = μ_{0} H$ .As the space inside the toroid is filled with aluminium the field becomes
$B = μ H = μ_{0} (1 + χ) H$
The increase in the field is $B - B_{0} = μ_{0} χ H$
The percentage increase is $\frac{B - B_{0}}{B} \times 100 = \frac{μ_{0} χ H}{μ_{0} H} \times 100 =$ $χ \times 100 = 2.1 \times 10^{- 3} (∵ χ = 2.1 \times 10^{- 5})$
By comparison, p=3

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Magnetic properties of matter

Remember concepts with our Masterclasses.

The percentage increase in the magnetic field B when the space within a current carrying toroid is filled with aluminium, with a susceptibility of 2.1×10-5, is 2.1 × 10-p. Find p.

In the absence of aluminium the magnetic field is B0=μ0H.As the space inside the toroid is filled with aluminium the field becomes B=μH=μ0(1+χ)HThe increase in the field is B-B0=μ0χHThe percentage increase is B-B0B×100=μ0χHμ0H×100 = χ×100=2.1×10-3 ∵χ=2.1×10-5By comparison, p=3

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The percentage increase in the magnetic field B when the space within a current carrying toroid is filled with aluminium, with a susceptibility of $2.1 \times 10^{- 5}$ , is $2.1 \times 10^{- p} .$ Find $p$ .