Two magnets are suspended by a given wire one by one. In order to deflect the first magnet through 45^ , the wire has to be twisted through 540^ whereas with the second magnet, the wire requires a twist of 360^ for the

Correct answer: A. 11 / 7 If C is torque per unit angular twist of the wire, then for a twist φ , =C φ=M B θ . In the 1^ st case, φ_1=540^ -45^ =495^ , θ_1=45^ In the 2^ nd case, φ_2=360^ -45^ =315^ , θ_2=45^ Dividing (i) by (ii), we get (M_1)/(M_2)=(495)/(315)=(11)/(7)

AIIMS 201826 May

Exam: AIIMS
Year: 2018
Date: 26 May
Subject: Physics
Chapter: Magnetic Effects of Current

AIIMS 2018 · 26 May · Physics · Magnetic Effects of Current

Two magnets are suspended by a given wire one by one. In order to...

Two magnets are suspended by a given wire one by one. In order to deflect the first magnet through

45^{\circ}

, the wire has to be twisted through

540^{\circ}

whereas with the second magnet, the wire requires a twist of

360^{\circ}

for the same deflection. Then the ratio of magnetic moments of the two magnets is

Options

A $11 / 7$
B $3 / 2$
C $4 / 3$
D $7 / 6$

Correct answer

A. $11 / 7$

Step-by-step solution

C

is torque per unit angular twist of the wire, then for a twist

\phi

\tau=C \phi=M B \sin \theta

. In the

1^{\text{st }}

case,

\phi_1=540^{\circ}-45^{\circ}=495^{\circ}, \theta_1=45^{\circ}

In the

2^{\text{nd }}

case,

\phi_2=360^{\circ}-45^{\circ}=315^{\circ}, \theta_2=45^{\circ}

Dividing (i) by (ii), we get

\frac{M_1}{M_2}=\frac{495}{315}=\frac{11}{7}

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