Question

A charge having q/m ewual to 10 8 /kg and with velocity 3×10^5 m/s enters into a uniform magnetic field B=0.3 T at an angle 30° with direction of field the radius of curvature will be?

Answer 1

Answer:

The radius of curvature will be 2 cm

Explanation:

According to the problem the charge has q/m = 10^8 C/kg

the velocity of the charge ,v = 3 x 10^5 m/s

The mentioned magnetic field ,B = 0.3 T

and the angle ,θ = 30°

We know the force = mv^2/r

  again force= qvBsinθ

qvBsinθ= mv^2/r [ where r = radius of curvature]

r= v/qBsinθ

 = 3 x 10'^5/10^8 0.3sin30°

 = 0.02 m

 = 2 cm