JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Magnetic Effects of Current

A particle of mass

M

and positive charge

Q

, moving with a constant velocity

u_{1} = 4 \hat{i} m s^{- 1}

, enters a region of uniform static magnetic field normal to the

x - y

plane. The region of the magnetic field extends from

x = 0

x = L

for all values of

y

. After passing through this region, the particle emerges on the other side after 10 milliseconds with a velocity

u_{2} = 2 (3 \hat{i} + \hat{j}) ms^{- 1}

. The correct statement(s) is (are)
(A) The direction of the magnetic field is

- z

direction.
(B) The direction of the magnetic field is

+ z

direction.
(C) The magnitude of the magnetic field

\frac{50 π M}{3 Q}

units.
(D) The magnitude of the magnetic field is

\frac{100 π M}{3 Q}

units.

Answer: (A), (C)

Solution

Initial velocity:

u_{1} = 4 \hat{i}

m/s

Final velocity:

u_{2} = 23 \hat{i} + 2 \hat{j}

m/s

The angle of deflection:

tan θ = \frac{2}{2 3} = \frac{1}{3}

θ = 30° = \frac{π}{6}

Since the particle deflects in the positive

y

direction and has positive charge, by the right-hand rule, the magnetic field must be in the

- z

direction. ✓ (A is correct)

Time in the field:

t = \frac{θ}{ω} = \frac{θ M}{QB}

10 \times 1 0^{- 3} = \frac{π M}{6 QB}

B = \frac{π M}{6 Q \times 10 \times 1 0 ^{- 3}} = \frac{π M \times 100}{6 Q} = \frac{50 π M}{3 Q}

✓ (C is correct)

Hence the correct answers are (A) and (C).

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · One or more correct options