A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque needed to maintain the needle in this position will be

Question

A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque needed to maintain the needle in this position will be (A) √ 3 W (B) W (C) (√ 3 /2)W (D) 2W

Tardigrade Admin · Accepted Answer

W = −MB(cos θ2 − cos θ1) Initially magnetic needle is parallel to a magnet field, therefore θ1 = 0, θ2 = 60° ∴ W = −MB(cos 60° − cos 0°) =MB e =MB sin 60° =ZW×√3/ 2=√3W

Q. A magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $6 0^{°}$ . The torque needed to maintain the needle in this position will be

$3 W$

$W$

$(3 /2) W$

$2 W$

$W = - MB (cos θ_{2} - cos θ_{1})$
Initially magnetic needle is parallel to a magnet field, therefore
$θ_{1} = 0,$
$θ_{2} = 6 0^{°}$
$∴ W = - MB (cos 6 0^{°} - cos 0^{°})$
$= MB$
$e = MB s in 6 0^{°} = Z W \times 3 /2 = 3 W$

Q. A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque needed to maintain the needle in this position will be

3​W

W

(3​/2)W

2W

W=−MB(cosθ2​−cosθ1​) Initially magnetic needle is parallel to a magnet field, therefore θ1​=0, θ2​=60° ∴W=−MB(cos60°−cos0°) =MB e=MBsin60°=ZW×3​/2=3​W

Q. A magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $6 0^{°}$ . The torque needed to maintain the needle in this position will be

$3 W$

$W$

$(3 /2) W$

$2 W$

$W = - MB (cos θ_{2} - cos θ_{1})$
Initially magnetic needle is parallel to a magnet field, therefore
$θ_{1} = 0,$
$θ_{2} = 6 0^{°}$
$∴ W = - MB (cos 6 0^{°} - cos 0^{°})$
$= MB$
$e = MB s in 6 0^{°} = Z W \times 3 /2 = 3 W$