A charged particle carrying charge `q=1muc` moves in uniform magnetic with velocity `v_1=10^6m//s` at angle `45^@` with `x`-axis in the `xy`-plane and experiences a force `F_1=5sqrt2mN` along the negative `z`-axis. When te same particle moves with velocity `v_2=10^6m//s` along the `z`-axis it experiences a force `F_2` in `y`-direction. Find a. the magnitude and direction of the magnetic field b. the magnetic of the force `F_2`.
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`F_2` is in `y`-direction when velocity is along `z`-axis. Therfore, magnetic field should be along `x`-axis. So let, `B=B_0hati` a. Given, `v_1=10^6/sqrt2hati+10^6/sqrt2hatj` and `F_1=-5sqrt2xx10^-3hatk` From the equation `F=q(vxxB)` We have `(-5sqrt2xx10^-3)hatk=(10^-6)[(10^6/sqrt2hati+10^6/sqrt2hatj)xx(B_0hati)]` `=-B_0/sqrt2hatk` `:. B_0/sqrt2=5sqrt2xx10^-3` or `B_0=10^-2T` Therefore, the magnetic field is `B=(10^-2hati)T` b. `F_2=B_0qv_2sin90^@` As the angle between `B` and `v` in this case is `90@` `:. F_2=(10^-2)(10^-6)(10^6)` `=10^-2N`
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