A uniform magnetic field of intensity 1T is applied in a circular region of radius 0.1 m , directed into the plane of paper . A charged particle of mass 5 x 10 ^-5 kg and charge q = 5 x 10 ^-4 C enters the field with vel. 1/root 3 m/s makin an angle of phi with a radial line of circular region in such a way that it passes thru center of applied field . The angle phi is:

(A) 60 degrees

(B) 30 degrees

(C) 45 degrees

(D) 90 degrees

To determine the angle phi at which the charged particle enters the magnetic field, we can use the principles of circular motion and the Lorentz force acting on the particle. Let's break down the problem step by step.

Understanding the Forces at Play

When a charged particle moves through a magnetic field, it experiences a magnetic force that acts perpendicular to both the velocity of the particle and the direction of the magnetic field. This force can cause the particle to move in a circular path if the conditions are right.

Key Variables

• Magnetic Field Intensity (B): 1 T (Tesla)
• Radius of the Circular Region (r): 0.1 m
• Mass of the Particle (m): 5 x 10^-5 kg
• Charge of the Particle (q): 5 x 10^-4 C
• Velocity (v): 1/√3 m/s

Applying the Formula for Circular Motion

The magnetic force acting on the particle can be expressed as:

F = qvB sin(θ)

Where θ is the angle between the velocity vector and the magnetic field. In our case, since the magnetic field is directed into the plane of the paper, we can simplify our analysis by considering the component of the velocity that is perpendicular to the magnetic field.

Finding the Radius of the Circular Path

The centripetal force required to keep the particle moving in a circular path is given by:

F_c = m(v²/r)

Setting the magnetic force equal to the centripetal force gives us:

qvB sin(θ) = m(v²/r)

Solving for the Angle

We can rearrange this equation to find sin(θ):

sin(θ) = (mv)/(qBr)

Now, substituting the known values:

• m = 5 x 10^-5 kg
• v = 1/√3 m/s
• q = 5 x 10^-4 C
• B = 1 T
• r = 0.1 m

Plugging in these values:

sin(θ) = (5 x 10^-5 kg * (1/√3) m/s) / (5 x 10^-4 C * 1 T * 0.1 m)

Calculating the right-hand side:

sin(θ) = (5 x 10^-5 / (5 x 10^-4 * 0.1)) * (1/√3)

sin(θ) = (5 x 10^-5 / 5 x 10^-5) * (1/√3)

sin(θ) = (1/√3)

Determining the Angle

Now, we can find θ by taking the inverse sine:

θ = sin^-1(1/√3)

Calculating this gives us:

θ ≈ 30 degrees

Final Answer

Thus, the angle phi at which the charged particle enters the magnetic field is 30 degrees, which corresponds to option (B).