Question

A charge having q/m equal to 10c / kg andwith velocity 3x10m/s enters into auniform magnetic field B = 0.3 tesla at anangle 30° with direction of field. Then radiusof curvature will be:(a) 0.01 cm (b) 0.5cm(c) 1 cm(d) 2cm​

Answer 1

Answer:

The answer will be 20 cm

Explanation:

According to the problem the charge of the particle, q = 10 C/kg

The velocity of the particle is v = 3 x 10 m/s

it enters the  magnetic field B = 0.3 Tesla

The particle enters into the magnetic field with the angle of 30°

Let the radius of the curvature = r

Therefore,

qvBsinθ = mv^2/r

r = mv/qBsinθ

  = 3 x 10/ 10 x 0.3 x sin30°

   = 3 x 10/ 10 x 0.3 x 1/2

    = 20 cm

Answer 2

The radius of curvature of the path is 2 cm.

Explanation:

It is given that,

The value of q/m is $10\ C/kg$

Velocity of the charged particle is $3\times 10^{-2}\ m/s$

The particle makes an angle of 30° with direction of field. We need to find the radius of curvature. The expression for the radius of circular path is given by :

$r=\dfrac{mv}{qB\sin \theta}$

Putting all the values in above formula we get :

$r=\dfrac{v}{(q/m)B\sin \theta}\\\\r=\dfrac{3\times 10^{-2}}{10\times 0.3\times \sin (30)}\\\\r=0.02\ m\\\\\text{or}\\\\r=2\ cm$

So, the radius of curvature of the path is 2 cm.

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