Question

A magnetized needle of magnetic moment 4.8 × 10–2 J T–1 is placed at 30° with the direction of uniform magnetic field of magnitude 3 × 10–2 T. Calculate the torque acting on the needle.

Answer 1

$\huge\underline\blue{\rm Answer:}$

$\large\red{\boxed{\sf Torque( \tau) =7.2×10^{-4}JT^{-1}}}$

$\huge\underline\blue{\rm Solution:}$

$\large\underline\pink{\sf Given: }$

• Magnetic Moment (M) = $\sf{4.8×10^{-2}JT^{-1}}$

• $\sf{\theta = 30°}$

• Magnetic Field (B) = $\sf{3×10^{-2}T}$

$\large\underline\pink{\sf To\:Find: }$

• Torque ($\sf{\tau}$)= ?

━━━━━━━━━━━━━━━━━━━━━━━━━

$\Large{♡}$$\large{\boxed{\sf Torque= MBsin\theta}}$

$\large\implies{\sf 4.8×10^{-2}×3×10^{-2}sin30°}$

$\large\implies{\sf 4.8×10^{-2}×3×10^{-2}×\frac12 }$

$\large\implies{\sf 7.2×10^{-4} }$

$\Huge\red{♡}$$\large\red{\boxed{\sf Torque (\tau) =7.2×10^{-4}JT^{-1}}}$