
If a bar magnet has its coercivity of $4000\,A{m^{ - 1}}$. It is desired to demagnetize by inserting it inside a solenoid $10\,cm$ long and having $500$ turns. The current which should be carried by solenoid is (consider an ideal solenoid).
A. $0.8{\mu _ \circ }A$
B. $0.4{\mu _ \circ }A$
C. $0.8A$
D. $\dfrac{{0.8}}{\mu }A$
Answer
412.8k+ views
Hint: The coercivity of a bar magnet is the magnetic field that will be required. The coercivity of a bar magnet is its resistance to becoming demagnetized. For this calculation we are using the coercivity formula, and we are converting given data into meters.
Complete step by step answer:
Given that, the coercivity of a bar magnet is $4000A{m^{ - 1}}$, $4 \times {10^3}$
Length of the solenoid is 10cm, $ = 10 \times {10^{ - 2}}$
From coercivity formula we have,
${H_c} = 4 \times {10^3} \times 10 \times {10^{ - 2}} \\
\Rightarrow {H_c} = 400A{m^{ - 1}} \\ $
Required current through solenoid is,
$i = \dfrac{{{H_c}}}{n}$
Already we have calculated the ${H_c}$ value,
Given number of turns is, $n=500$
By substituting the values in current i, then we get,
$i = \dfrac{{400}}{{500}} \\
\therefore i = 0.8A \\ $
Hence, the correct option is C.
Additional information:
$H$ field is required to reduce the magnetic flux ($B$) (B inside the material is zero) to zero is called normal coercivity. Whereas, $H$ field is required to reduce the magnetization is called intrinsic coercivity. In ferromagnetic material the intensity of applied magnetic field, which is required to demagnetize the material is called coercivity. Demagnetizing field is applied to the opposite of the original saturating field.
Note: From the given data we have calculated the coercivity of solenoid from using the formula of coercivity. The distance between intrinsic and normal coercivity is negotiable.In Normal coercivity, $H$ field is required to reduce the magnetic flux to zero. In Intrinsic coercivity, $H$ field is required to reduce the magnetization to zero. In Remanence coercivity, $H$ field is required to reduce the remanence to zero, and that $H$ is finally returned to the zero.
Complete step by step answer:
Given that, the coercivity of a bar magnet is $4000A{m^{ - 1}}$, $4 \times {10^3}$
Length of the solenoid is 10cm, $ = 10 \times {10^{ - 2}}$
From coercivity formula we have,
${H_c} = 4 \times {10^3} \times 10 \times {10^{ - 2}} \\
\Rightarrow {H_c} = 400A{m^{ - 1}} \\ $
Required current through solenoid is,
$i = \dfrac{{{H_c}}}{n}$
Already we have calculated the ${H_c}$ value,
Given number of turns is, $n=500$
By substituting the values in current i, then we get,
$i = \dfrac{{400}}{{500}} \\
\therefore i = 0.8A \\ $
Hence, the correct option is C.
Additional information:
$H$ field is required to reduce the magnetic flux ($B$) (B inside the material is zero) to zero is called normal coercivity. Whereas, $H$ field is required to reduce the magnetization is called intrinsic coercivity. In ferromagnetic material the intensity of applied magnetic field, which is required to demagnetize the material is called coercivity. Demagnetizing field is applied to the opposite of the original saturating field.
Note: From the given data we have calculated the coercivity of solenoid from using the formula of coercivity. The distance between intrinsic and normal coercivity is negotiable.In Normal coercivity, $H$ field is required to reduce the magnetic flux to zero. In Intrinsic coercivity, $H$ field is required to reduce the magnetization to zero. In Remanence coercivity, $H$ field is required to reduce the remanence to zero, and that $H$ is finally returned to the zero.
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