Question

A proton is projected with a velocity 107 ms-1 at right angles to a uniform magnetic field of induction 100 mT. The time (in seconds) taken by the proton to traverse 90o arc is(Mass of proton = 1.65 × 10-27 kg and charge of proton = 1.6 × 10-19 C)

• A. (A) 0.81 × 10-7
• B. (B) 1.62 × 10-7
• C. (C) 2.43 × 10-7
• D. (D) 3.24 × 10-7

Answer 1

Answer: (B)

(B) 1.62 × 10-7

Answer 2

Explanation:
When proton enters in a uniform magnetic field at right angle then it moves on a circular path. In this case velocity of proton.v=Bqrmwhere r= radius of circular path Distance covered by proton to traverse 90∘Time taken by proton to cover distance dt=πr/2v=πr2vt=π2rBqr/mt=πm2BqPutting m=1.65×10−27kgB=100mT=100×10−3T,q=1.6×10−19Ct=3.14×1.65×10−272×100×10−3×1.6×10−19t=1.62×10−7sec