The coercivity of a magnet is 5 × 103 A/m. The amount of current required to be passed in a solenoid of length 30 cm and the number of turns 150, so that the magnet gets demagnetised when inside the solenoid is ………A.
Correct Answer: 10
Approach Solution - 1
To determine the current required to demagnetize the magnet with coercivity \(H_c = 5 \times 10^3\) A/m, we use the formula for the magnetic field inside a solenoid: \(B = \mu_0 nI\), where \(B\) is the magnetic field, \(n\) is the number of turns per unit length, and \(I\) is the current.
Step 1: Find the number of turns per meter \(n\).
Given: Length of solenoid \(l = 30\) cm = 0.3 m, Number of turns \(N = 150\).
Turns per unit length \(n = \frac{N}{l} = \frac{150}{0.3} = 500\) turns/m.
Step 2: Use the coercivity to find the required current \(I\).
The coercivity is effectively the magnetic field needed to demagnetize the magnet: \(H_c = nI\). Solving for \(I\), we have:
\(I = \frac{H_c}{n} = \frac{5 \times 10^3}{500} = 10\) A.
Step 3: Confirm the current is within the range.
The computed current \(I = 10\) A is within the provided range of 10 to 10. Thus, the result is valid.
Conclusion: The required current to demagnetize the magnet is 10 A.
Approach Solution -2
The coercivity of the magnet is given by:
\[H_c = \mu_0 \frac{ni}{\mu_0},\]
where:
\[H_c = 5 \times 10^3 \, \text{A/m}, \quad n = \frac{\text{Number of turns}}{\text{Length of solenoid}} = \frac{150}{0.3} = 500 \, \text{turns/m}.\]
Substitute into the formula:
\[5 \times 10^3 = 500 \times i.\]
Solve for \(i\):
\[i = \frac{5 \times 10^3}{500} = 10 \, \text{A}.\]
Thus, the current required is:
\[i = 10 \, \text{A}.\]
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