Question

Q4. A bar magnet when suspended horizontally and perpendicular to earth's field
experiences a torque of 3 x 10 Nm. What is the magnetic moment of the magnet ? If horizontal
component of earth's magnetic field at that place is 04 *10tesla.​

Answer 1

Answer:

Explanation:

Here, θ=90∘, τ=3×10−4Nm,

M=?, B=0⋅4×10−4tesla

As τ=MBsinθ

∴M=τBsinθ=3×10−40⋅4×10−4sin90∘=7⋅5JT−1

Answer 2

Explanation:

• So there is a bar magnet perpendicular to the earth’s magnetic field.
• So vector M is perpendicular to magnetic field B
•    Now torque = vector M X B
•                         = MB sin theta
•                    So theta = 90 degree
•               Or torque = MB sin 90  
•                Or torque = MB (sin 90 = 1)
• Now we need to find magnetic moment.
•                       So M = torque / B
•   Substituting we get M = 3 x 10^-4 / 0.4 x 10^-4
•                            Or M = 30 / 4
•                              Or M = 7.5 J/T

Reference link will be

https://brainly.in/question/11357748