Question

A particle having charge 6C enters perpendicularly in a uniform magnetic field of strength 100T with a momentum of
36kg m/s. The radius of the circular path of the particle is
$[Take{\mu }_0=4\times {10}^{–7}Tm/A]$

• A. 60 cm
• B. 6 m
• C. 6 cm
• D. 6 mm

Answer 1

The correct option is C 6 cm

Charge, q = 6 C
Magnetic field, B = 100 T
Momentum of the charged particle, mv = 36 kg m/s
Angle between velocity and magnetic field $\theta ={90}^{\circ }$
Let r be the radius of the circular path.
$Now,\frac{m{v}^2}{r}=q\overline{v}\times \overline{B}=qvBsin{90}^{\circ }=qvb$
$\Rightarrow r=\frac{mv}{qB}=\frac{36}{6\times 100}$
= 0.06 m
= 6 cm
Hence, the correct answer is option (3).