If the 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 it through an angle of 60°, then the value of n is

Question

If the 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 it through an angle of 60°, then the value of n is (A) 1 (B) 2 (C) (1/2) (D) (1/4)

Tardigrade Admin · Accepted Answer

We have, W=-M B( cos θ2- cos θ1) So, W1=-M B( cos 90°- cos 0°)=M B and W2=-M B( cos 60°- cos 0°)=(1/2) M B As W1=n W2 ∴ n=(W1/W2)=(M B/(1)2 M B)=2

Q. If the work done in turning a magnet of magnetic moment $M$ by an angle of 90 $^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of 60 $^{\circ}$ , then the value of $n$ is

1

2

$\frac{1}{2}$

$\frac{1}{4}$

We have, $W = - MB (cos θ_{2} - cos θ_{1})$
So, $W_{1} = - MB (cos 9 0^{\circ} - cos 0^{\circ}) = MB$
and $W_{2} = - MB (cos 6 0^{\circ} - cos 0^{\circ}) = \frac{1}{2} MB$
As $W_{1} = n W_{2}$
$∴ n = \frac{W _{1}}{W _{2}} = \frac{MB}{\frac{1}{2} MB} = 2$

Q. If the 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 it through an angle of 60∘, then the value of n is

1

2

21​

41​

We have, W=−MB(cosθ2​−cosθ1​) So, W1​=−MB(cos90∘−cos0∘)=MB and W2​=−MB(cos60∘−cos0∘)=21​MB As W1​=nW2​ ∴n=W2​W1​​=21​MBMB​=2

Q. If the work done in turning a magnet of magnetic moment $M$ by an angle of 90 $^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of 60 $^{\circ}$ , then the value of $n$ is

$\frac{1}{2}$

$\frac{1}{4}$

We have, $W = - MB (cos θ_{2} - cos θ_{1})$
So, $W_{1} = - MB (cos 9 0^{\circ} - cos 0^{\circ}) = MB$
and $W_{2} = - MB (cos 6 0^{\circ} - cos 0^{\circ}) = \frac{1}{2} MB$
As $W_{1} = n W_{2}$
$∴ n = \frac{W _{1}}{W _{2}} = \frac{MB}{\frac{1}{2} MB} = 2$