Question

A bar magnet of moment 7.5 A-m^2 experience a torque of 1.5×10^-4 N-m, when placed inclined at 30° in a uniform magnetic field. Find the magnetic induction ofa field.

Answer 1

Answer:

Magnetic Induction = $4*10^{-5}\:Wb/m^2$

Step by step explanation:

It is given that:

• Magnetic moment 'M'=$7.5\:Am^2$
• Angle with field, $\theta =30^o$
• Torque $T = 1.5*10 ^ { - 4 }$ Nm

It is required to calculate the magnetic induction 'B'.

Magnetic Induction is the production of an electromotive force across an electrical conductor in a changing magnetic field.

$T=MBsin\theta\\ \\B=\frac{T}{Msin\theta}\\ \\B=\frac{1.5*10^{-4}}{7.5sin(30^o)}\\ \\B=4*10^{-5}\:Wb/m^2$