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Work done required to turn a magnet by 9...

Work done required to turn a magnet by `90^@` from magnetic meridian is n times to that when it is rotated by `60^@` angle. What is the value of n?

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To solve the problem, we need to calculate the work done in turning a magnet from the magnetic meridian by two different angles: 90 degrees and 60 degrees. We will use the formula for potential energy of a magnetic dipole in a magnetic field, which is given by: \[ U = -mB \cos \theta \] Where: - \( U \) is the potential energy, - \( m \) is the magnetic moment, - \( B \) is the magnetic field strength, - \( \theta \) is the angle between the magnetic moment and the magnetic field. ### Step 1: Calculate the work done for 90 degrees rotation 1. **Initial angle (\( \theta_1 \))**: 0 degrees (magnet aligned with the field) 2. **Final angle (\( \theta_2 \))**: 90 degrees (magnet perpendicular to the field) Using the formula for work done \( W \): \[ W_1 = U(\theta_1) - U(\theta_2) \] \[ W_1 = (-mB \cos 0^\circ) - (-mB \cos 90^\circ) \] \[ W_1 = -mB(1) - (-mB(0)) \] \[ W_1 = -mB + 0 \] \[ W_1 = mB \] ### Step 2: Calculate the work done for 60 degrees rotation 1. **Initial angle (\( \theta_1 \))**: 0 degrees 2. **Final angle (\( \theta_2 \))**: 60 degrees Using the same work done formula: \[ W_2 = U(\theta_1) - U(\theta_2) \] \[ W_2 = (-mB \cos 0^\circ) - (-mB \cos 60^\circ) \] \[ W_2 = -mB(1) - (-mB(1/2)) \] \[ W_2 = -mB + \frac{mB}{2} \] \[ W_2 = -\frac{mB}{2} \] ### Step 3: Relate the two work done expressions According to the problem, the work done for the 90 degrees rotation is \( n \) times the work done for the 60 degrees rotation: \[ W_1 = n \cdot W_2 \] Substituting the values we calculated: \[ mB = n \left(-\frac{mB}{2}\right) \] ### Step 4: Solve for \( n \) Rearranging the equation: \[ mB = -\frac{n \cdot mB}{2} \] Dividing both sides by \( mB \) (assuming \( mB \neq 0 \)): \[ 1 = -\frac{n}{2} \] Multiplying both sides by -2: \[ n = -2 \] However, since we are looking for the absolute value of \( n \) in the context of work done, we take: \[ n = 2 \] ### Final Answer The value of \( n \) is \( 2 \). ---

To solve the problem, we need to calculate the work done in turning a magnet from the magnetic meridian by two different angles: 90 degrees and 60 degrees. We will use the formula for potential energy of a magnetic dipole in a magnetic field, which is given by: \[ U = -mB \cos \theta \] Where: - \( U \) is the potential energy, - \( m \) is the magnetic moment, - \( B \) is the magnetic field strength, ...
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