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AIEEE Corner Subjective Questions (Level-1) ® ^ ^ ^ 7. Let B = Bx i + By j + Bz k
1. Positive. By Flemings left hand rule. 2. Fm = evB sin q Fe Þ v= eB sin q =
® ® ® (a) F = q( v ´ B ) ^
1.6 ´ 10-19 ´ 3.5 ´ 10-3 ´ sin 60°
^
Þ Bx = - 0.175 T, Bz = - 0.256 T (b) Cannot be determined by this information. ® ® ® (c) As F = q ( v ´ B )
3. Fm = qvB sin q = (2 ´ 1.6 ´ 10-19 ) ´ 105 ´ 0.8 ´ 1 = 2.56 ´ 10-14 N ® ® ® 4. (a) Fm = e ( v ´ B ) ^
® ® F ^B ^
= - 1.6 ´ 10-19 [(2.0 ´ 106 ) i + (3.0 ´ 106 ) j] ^
^
´ (0.03 i + 0.15 j)
®® Hence, B × F = 0 ® ^ 8. B = B i ® ^ (a) v = v j
^
= - (6.24 ´ 10-4 N) k
® ® ® ^ F = q ( v ´ B ) = - qvB k
® ® ® ^ (b) = Fm e ( v ´ B ) = - (6.24 ´ 10-4 N) k ® ^ (b) v = v j
® ® ® 5. Fm = e ( v ´ B ) ^
^
® ® ® ^ F = q ( v ´ B ) = qvB j
^
(6.4 ´ 10-19 ) k = - 1.6 ´ 10-19[(2 i + 4 j) ^
^
´ ( Bx i + 3 Bx j)] ^
® ^ (c) v = - v i ® ® ® F = q( v ´ B ) = 0
^
6.4 ´ 10-19 k = - 1.6 ´ 10-19[2Bx k ] 6.4 ´ 10-19 - 3.2 ´ 10-19
= - 2.0 T
6. (a) As magnetic force always acts perpendicular to magnetic field, magnetic field must be along x-axis. F1 = qv1B sin q1 5 2 ´ 10-3 F1 Þ B= = qv1B sin q1 1 ´ 10-6 ´ 106 ´ 1 2 Þ or
B = 10-3 T ® ^ B = (10-3 T ) i
(b) F2 = qv2 B sin q2 = 1 ´ 10-6 ´ 106 ´ 10-3 ´ sin 90° = 10-3 N F2 = 1 mN
^
= - 7.8 ´ 10-6 ´ 3.8 ´ 10 3( Bz i - Bx k )
= 9.46 ´ 106 m / s
Bx =
^
7.6 ´ 10-3 i - 5.2 ´ 10-3 k
4.6 ´ 10-15
® ^ ^ (d) v = v cos 45° i - v cos 45° k ® ® ® qvB ^ F = q( v ´ B ) = j 2 ® ^ ^ (e) v = v cos 45° j - v cos 45° k ® ® ® qvB ^ ^ F = q( v ´ B ) = (- j - k) 2 qvB ^ ^ =( j + k) 2 2m k 2 m eV mv 9. r = = = qB eB eB B=
Þ =
2mV r e
2 ´ 9.1 ´ 10-31 ´ 2 ´ 103 1.6 ´ 10-19
´ 0.180