Question

A conductor of length 25 cm is placed (i) parallel (ii) perpendicular (iii) inclined at an angle 30°, to a uniform magnetic field of induction 2 T. If 1 C of charge passes through it in 5 s, calculate the force experienced by the conductor in each case.

Answer 1

Data : l = 25 cm = 0.25 m, 

θ₁ = 0°, 

θ₂ = 90°, 

θ₃ = 30°, B = 2 T, 

q = 1 C, f = 5 s

The current in the conductor,

I = \(\cfrac qt\) = \(\cfrac{1C}{5s}\) = 0.2 A

The magnitude of the force on the conductor,

F = I\(\vec l\) x \(\vec B\)| = IlB sin θ

(i) θ₁ = 0° A 

∴ sin θ = 0 

∴ F = 0 N 

(ii) θ₂ = 9O° 

∴ sin θ = 1 

∴ F = IlB = (0.2 A)(0.25 m)(2 T)

= 0.1 N 

(iii) θ₃ = 30° 

∴ sin θ₃ = 0.5

∴ F = IlB sin θ₃ = (0.1 N) (0.5) = 0.05 N

The direction of \(\vec F\) in each case is given by the cross product rule.