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A charged particle carrying charge `q=10 muC` moves with velocity `v_1=10^6 ,s^-1` at angle `45^@` with x-axis in the xy plane and experience a force `F_1=5sqrt2 mN` along the negative z-axis. When the same particle moves with velocity `v_2=10^6 ms^-1` along the z-axis, it experiences a force `F_2` in y-direction. Find the magnetic field `vecB`.A. `(10^-3T)(hati+hatj)`B. `(2xx10^-3T)hati`C. `(10^-3T)hati`D. `(2xx10^-3T)(hati+hatj)` |
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Answer» Correct Answer - c For the first case: `vecF=qvecvxxvecB` `implies -5sqrt2xx10^-3hatk` `=10^-5xx(10^6)/(sqrt2) (hati+hatj)xx(B_xhati+B_yhatj+B_zhatk)` `=(10/(sqrt2))[B_zhati-B_zhatj+(B_y+B_x)hatk]` `implies B_z=0, B_y-B_x=-10^-3T....(i)` Similarly, for the second case: `F_2hatj=(10^-5)(10^6hatk)xx[(B_xhati+B_yhatj+B_zhatk)]` `F_2hatj=10(B_xhatj-B_yhati)` `F_2=10B_x, B_y=0....(ii)` Using eqs. (i) and (ii), we get `B_x=10^-3T` Thus, `vecB=(10^-3T) hati` Also, `F_2=10B_x=10^-2N`. |
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