Class 12 Physics
Moving Charges and Magnetism
Motion of a Charge in a Magnetic Field:
- For a charge q moving with velocity v in the presence of magnetic field B, force FB is given by:
FB = q(vxB) = qvB(sinθ)ȓ
- Case-1 (charge moving perpendicular to the magnetic field):When motion of charge v and the magnetic field B are at right angle (90°) to each other, the charge will follow a circular path with forceFB always acting towards the center (Centripetal force) and the velocityvacting tangentially to the circle
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Direction of magnetic field (B) is normal to the plane of paper and inwards |
- Case-2 (charge moving at an angle θ to the plane of magnetic field):When a charge in motion moves such that the angle between the moving charge and the plane of magnetic field is θ, then the velocity (v) of charge has 2 components, one component along the direction of magnetic field (vcosθ), and the another perpendicular to the magnetic field (vsinθ)
- The component of velocity (vcosθ) along the magnetic field direction, is responsible for the uniform motion of charge along the direction of magnetic field
- The component (vsinθ) perpendicular to the magnetic field will make the charge to follow a circular pathdue to centripetal force FB (just as in case-1)
- These 2 mutually independent motions will cause the charge to move in a helical path as shown below
- To find the radius r of the helix, we can use the centripetal force FC which is provided the force due to magnetic field FB:
FC = FB
mv2/r = qvBsinθ
mv/r = qBsinθ
∴ r = mv/(qBsinθ)
- To find the angular frequency wof the motion of charge:
v = w x r
w = v/r = qB(sinθ)/m
- To find the time T of each revolution:
T = 2π/w = 2πm/(qBsinθ)
- NOTE: The above equations prove that frequency(w) or time period(T) doesn’t depend on the velocity v (or energy) of charge particle
- To find the pitchp of helix, we can use the component of charge velocity along the magnetic field (vcosθ), and the time for each revolution T:
p = v(cosθ) x T = vcosθ x 2πm/(qBsinθ)
∴ p = 2πmv(tanθ)/qB
Numerical Problem:
1) A uniform magnetic field of 6.5G (1G = 10-4T) is maintained. An electron is shot normal to the field with velocity of 4.8×106m/s. Will the motion of electron in the magnetic field be circular, if so, why?
Find the radius and kinetic energy of the electron.
Solution: We already know now that the force of moving charge in an external electric field is:
The direction of force(cross product of v and B) will always be normal to the plane of magnetic field lines, and also normal to the motion of charge (always). And the in the uniform circular motion, centripetal force is also always normal to the motion of object. Hence, the charge will execute a circular motion with force acting as centripetal force.
Radius of circular orbit will be given by: r = mv/(qBsinθ)
m = 9.1×10-31kg, q = 1.6×10-19C, v = 4.8×106m/s, B = 6.5×10-4T, θ = 90°
∴r = (9.1×10-31×4.8×106)/(1.6×10-19×6.5×10-4) = 0.042m = 4.2cm (ans)
Kinetic energy of electron will be:
K = ½ mv2 = (9.1×10-31×(4.8×106)2)/2 = 1.05×10-17J (ans)
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