The percentage increase in magnetic field B when the space within a current carrying toroid is filled with aluminium $(X = 2.1 \times 10^{- 5})$ is

0.2 %

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2 \times 10^{- 3} %

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2 \times 10^{- 2} %

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2 \times 10^{- 4} %

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Solution

The correct option is A $2 \times 10^{- 3} %$
Here, $X = 2.1 \times 10^{- 5}$
From $B = μ_{0} (H + I) = μ_{0} H (1 + \frac{I}{H}) = B_{0} (1 + X)$

$B - B_{0} = B_{0} X$

$\frac{B - B_{0}}{B_{0}} = X \Rightarrow \frac{(B - B_{0})}{B_{0}} \times 100 = 100 X = 100 \times 2.1 \times 10^{- 5} = 2.1 \times 10^{- 3} %$

Hence, option $(B)$ is correct.

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The percentage increase in magnetic field B when the space within a current carrying toroid is filled with aluminium (X=2.1×10−5) is

The correct option is A 2×10−3%Here, X=2.1×10−5From B=μ0(H+I)=μ0H(1+IH)=B0(1+X)B−B0=B0XB−B0B0=X⇒(B−B0)B0×100=100X=100×2.1×10−5=2.1×10−3%Hence, option (B) is correct.

The percentage increase in magnetic field B when the space within a current carrying toroid is filled with aluminium $(X = 2.1 \times 10^{- 5})$ is