Work done in turning a magnet of magnetic moment M by an angle of 90° from the magnetic meridian is n times the corresponding work done to turn through an angle of 60°, where n is

Question

Work done in turning a magnet of magnetic moment M by an angle of 90° from the magnetic meridian is n times the corresponding work done to turn through an angle of 60°, where n is (A) 1/2 (B) 2 (C) 1/4 (D) 1

Tardigrade Admin · Accepted Answer

W1=-M B( cos 90°- cos 0°)=M B W2=-M B( cos 60°- cos 0°) =-M B((1/2)-1)=(1/2) M B Which is =(1/2) W1 As, W1=n W2 ∴ n=2

Q. Work done in turning a magnet of magnetic moment $M$ by an angle of $9 0^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn through an angle of $6 0^{\circ}$ , where $n$ is

1/2

2

1/4

1

$W_{1} = - MB (cos 9 0^{\circ} - cos 0^{\circ}) = MB$
$W_{2} = - MB (cos 6 0^{\circ} - cos 0^{\circ})$
$= - MB (\frac{1}{2} - 1) = \frac{1}{2} MB$
Which is $= \frac{1}{2} W_{1}$
As, $W_{1} = n W_{2} ∴ n = 2$

Q. Work done in turning a magnet of magnetic moment M by an angle of 90∘ from the magnetic meridian is n times the corresponding work done to turn through an angle of 60∘, where n is

1/2

2

1/4

1

W1​=−MB(cos90∘−cos0∘)=MB W2​=−MB(cos60∘−cos0∘) =−MB(21​−1)=21​MB Which is =21​W1​ As, W1​=nW2​∴n=2

Q. Work done in turning a magnet of magnetic moment $M$ by an angle of $9 0^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn through an angle of $6 0^{\circ}$ , where $n$ is

$W_{1} = - MB (cos 9 0^{\circ} - cos 0^{\circ}) = MB$
$W_{2} = - MB (cos 6 0^{\circ} - cos 0^{\circ})$
$= - MB (\frac{1}{2} - 1) = \frac{1}{2} MB$
Which is $= \frac{1}{2} W_{1}$
As, $W_{1} = n W_{2} ∴ n = 2$