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  • A bar magnet having a magnetic moment of <img alt="\mathrm{2.0\times 10^{5}\, JT^{-1}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B2.0%5Ctimes%2010%5E%7B5%7D%5C%2C%20JT%5

A bar magnet having a magnetic moment of \mathrm{2.0\times 10^{5}\, JT^{-1}}, is placed along the direction of uniform magnetic field of magnitude \mathrm{B= 14\times 10^{-5}T}. The work done in rotating the magnet slowly through \mathrm{60^{\circ}} from the direction of field is :

Option: 1

\mathrm{14\, J}


Option: 2

\mathrm{8.4\, J}


Option: 3

\mathrm{4\, J}


Option: 4

\mathrm{1.4\, J}


Answers (1)

best_answer

\mathrm{M =2 \times 10^{5} }
\mathrm{B =14 \times 10^{-5} T }
\mathrm{\theta_{1}=0^{\circ} \quad(\bar{M}\, \| \, \bar{B} \text { direction of field }) }
\mathrm{\theta_{2}= 60^{\circ}}

\mathrm{w_{\text {ext }}=\Delta U=U_{f}-U_{i}}
            \mathrm{=\left(-M B \cos \theta_{2}\right)-\left(-M B \cos \theta_{1}\right) }
             \mathrm{=\left(-M B \cos 60^{\circ}\right)-\left(-M B \cos 0^{\circ}\right)} 
              \mathrm{=\frac{-M B}{2}+M B}
               \mathrm{=\frac{M B}{2}} 

\mathrm{W_{\text {ext }}=\frac{2 \times 10^{5} \times 14 \times 10^{-5}}{2}=14 \mathrm{~J} }

                                      

 

Posted by

vishal kumar

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