A proton of velocity (3 i +2 j ) m s ⁻¹ enters a field of magnetic induction (2 j +3 k ) tesla. The acceleration produced in the proton in m s ⁻² is (Specific charge of proton =0.96 10^8 C kg ⁻¹ )

Step-by-step solution

Here, v =3 i +2 j ~m ~s ⁻¹ B =2 j +3 k ~T Specific charge of proton = e m =0.96 10^8 C kg ⁻¹ Force on a proton in a uniform magnetic field is aligned & F =e( v B ) & =e[(3 i +2 j ) (2 j +3 k )] & =e(6 k -9 j +6 i )=e(6 i -9 j +6 k ) N aligned Acceleration of the proton, aligned & a = F m = e(6 i -9 j +6 k ) m & =0.96 10^8(6 i -9 j +6 k ) m s ⁻² & =2.88 10^8(2 i -3 j +2 k ) m s ⁻² aligned