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A charged particle going around in a circle can be considered to be a current loop. A particle of mass m carrying charge q is moving in a plane with speed v under the influence of magnetic field \overrightarrow{B}. The magnetic moment of this moving particle:  
Option: 1 -\frac{mv^{2}\overrightarrow{B}}{2B^{2}}
 
Option: 2 \frac{mv^{2}\overrightarrow{B}}{2\pi B^{2}}
Option: 3 \frac{mv^{2}\overrightarrow{B}}{B^{2}}  
Option: 4 \frac{mv^{2}\overrightarrow{B}}{2B^{2}}

Answers (1)

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\begin{aligned} &\text { Magnetic moment }\\ &\mathrm{M}=\mathrm{iA}\\ &M=\left(\frac{q}{T}\right) \times \pi r^{2}=\frac{q \pi r^{2}}{\left(\frac{2 \pi r}{v}\right)}=\frac{q v r}{2}\\ &M=\frac{q v}{2} \times \frac{v m}{q B}\\ &\mathrm{M}=\frac{\mathrm{mv}^{2}}{2 \mathrm{~B}} \end{aligned}

As we can see from the figure, direction of magnetic moment (M) is opposite to magnetic field.

\begin{array}{l} \overrightarrow{\mathrm{M}}=-\frac{\mathrm{mv}^{2}}{2 \mathrm{~B}} \hat{\mathrm{B}} \\ \\ =-\frac{\mathrm{mv}^{2}}{2 \mathrm{~B}^{2}} \overrightarrow{\mathrm{B}} \end{array}

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Deependra Verma

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