• # question_answer A particle of charge Q and of negligible initial speed is accelerated through a potential difference of U. The particle reaches a region of uniform magnetic field of induction B, where it undergoes circular motion. If potential difference is doubled and B is also doubled then magnetic moment of the circular current due to circular motion of charge Q will become - A) Double              B) half C) four times                      D) remain same

Correct Answer: D

Solution :

[D]  K.E. of charge Q accelerated with potential difference of U is given by  K.E. = QU Magnetic moment $=i\,\,\times \,\,Area$$=\frac{Q}{T}\,\,\times \,\,\pi {{R}^{2}}$ Where T is time period of circular motion of charge Q and R is radius of circular motion $\therefore T=\frac{2\pi m}{QB}$ $R=\sqrt{\frac{2mKE}{QB}}=\sqrt{\frac{2mU}{QB}}$ Magnetic moment $=\frac{{{Q}^{2}}\,\,\times \,\,B}{2\pi m}\times \frac{\pi \,\,\times \,\,2m\,\,\times \,\,U}{{{Q}^{2}}{{B}^{2}}}$ Magnetic Moment $\propto \frac{U}{B},$ so if U and B both are double magnetic moment remain unchanged.

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