# A wire loop carrying a current I is placed in the x – y plane as shown in fig. If a particle with charge + Q and mass m is placed at the centre P and given a velocity  along NP (see figure), find its instantaneous acceleration. If an external uniform magnetic induction field  = B  is applied, find the force and the torque acting on the loop due to this field.

Navjyot Kalra
9 years ago
Hello Student,
The magnetic field at the centre P due to current in wire NM is
B1 = μ­0 / 4 π I / r[sin 60° + sin 60°]
B1 μ­0 / 4 π I / a/ 2 x [ √3 / 2 + √ 3 / 2], B1 = μ0 / 4π 2I√3 / a
directed away from the reader perpendicular to the plane of paper.
sin 30° = r / a ⇒ r = a/ 2
cos 30° = MS / a
∴ MS = √3a / 2 ∴ MN = √3a
The magnetic field at the centre P due to current in are MN is
B2 = μ0 / 2π 2πI / a (θ / 2π) = μ0 / 4πI / a (2π / 3 / 2π ) = μ0 / 4π0 2πI / 3a
Directed towards the reader perpendicular to the plane of paper
The net magnetic field
B = B1 – B2 = μ0 / 4π 2√3I / a – μ0 / 4π 2πI / 3a
= μ0 2I / 4πa [ √ 3 – π / 3π] = μ0 / 4π x 2I / a x (0.68)
(directed away from the reader perpendicular to the plane of paper)
The force acting on the charged particle Q when it has a velocity v and is instantaneously at the centre is
F = QvB sin θ = QvB sin 90° = QvB
The acceleration produced
A = F / M = QvB / m = Qv / m [ μ0 / 4π x 2I / a x (0.68)]
A = 0.11 π0 IQv / ma
The direction of acceleration is given by the vector product x or by applying Fleming’s left hand rule
∴ ∠ MPR = = 120° - 90° = 30°
Since, ∠ MPQ = 60
∴ ∠RPQ = 30°
i.e., the acceleration vector makes an angle of 30° with the negative x – axis.
ALTERNATE SOLUTION
$\underset{B}{\rightarrow}$1 = μ0 / 4π 2I √ / a (- $\underset{k}{\rightarrow}$);$\underset{B}{\rightarrow}$2 = μ0 / 4π 2πI / 3a$\underset{k}{\rightarrow}$
KEY CONCEPT :T he torque acting on the loop in the magnetic field is given by
The force acting on the loop is zero.
Thanks
Navjot Kalra