46. A particle having charge q moves with a velocity through a region in which both an electric field and a magnetic field are present .The force on the particle is

1. 1. q^→_E + q(^→_B × ^→_v)

   
   
   

   2. q^→_E . (^→_B × ^→_v)

   
   
   

   3. q^→_v + q(^→_E × ^→_B)

   
   
   

   4. q^→_E + q(^→_v × ^→_B)

   
   
   

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1. View Hint View Answer Discuss in Forum

   ​Force due to electric field = q^→_E ​
   Force due to magnetic field = ​q(^→_v × ^→_B)
   Net force experienced = q^→_E + q(^→_v × ^→_B)
   

   Correct Option: D

   ​Force due to electric field = q^→_E ​
   Force due to magnetic field = ​q(^→_v × ^→_B)
   Net force experienced = q^→_E + q(^→_v × ^→_B)
   

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47. Two long parallel wires are at a distance of 1 metre. Both of them carry one ampere of current. The force of attraction per unit length between the two wires is​​

1. 1. 2 × 10^–7 N/m ​

   
   
   

   2. 2 × 10^–8 N/m

   
   
   

   3. 5 × 10^–8 N/m

   
   
   

   4. 10^–7 N/m

   
   
   

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1. View Hint View Answer Discuss in Forum

   F =	μ0	×	2i1i2I	= 10-7 ×	2 × 1 × 1 × 1	= 2 × 10-7 N/m.
   	4π		r		1

   
   ​[This relates to the definition of ampere]

   Correct Option: A

   F =	μ0	×	2i1i2I	= 10-7 ×	2 × 1 × 1 × 1	= 2 × 10-7 N/m.
   	4π		r		1

   
   ​[This relates to the definition of ampere]

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48. A coil carrying electric current is placed in uniform magnetic field, then

1. 1. torque is formed ​

   
   
   

   2. e.m.f is induced

   
   
   

   3. ​both (a) and (b) are correct

   
   
   

   4. none of the above

   
   
   

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1. View Hint View Answer Discuss in Forum

   A current carrying coil has magnetic dipole moment. Hence, a torque →pm acts →B on it in magnetic field.

   Correct Option: A

   A current carrying coil has magnetic dipole moment. Hence, a torque →pm acts →B on it in magnetic field.

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49. A straight wire of length 0.5 metre and carrying a current of 1.2 ampere is placed in uniform magnetic field of induction 2 tesla. The magnetic field is perpendicular to the length of the wire. The force on the wire is

1. 1. 2.4 N ​

   
   
   

   2. 1.2 N

   
   
   

   3. 3.0 N

   
   
   

   4. 2.0 N

   
   
   

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1. View Hint View Answer Discuss in Forum

   F = Bi ℓ = 2 ×1.2 × 0.5 = 1.2 N

   Correct Option: B

   F = Bi ℓ = 2 ×1.2 × 0.5 = 1.2 N

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50. In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be :

1. 1. 1	G
      499

   
   
   

   2. 499	G
      500

   
   
   

   3. 1	G
      500

   
   
   

   4. 500	G
      499

   
   
   

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1. View Hint View Answer Discuss in Forum

   As 0.2% of main current passes through the galvanometer hence 998/1000 current through the shunt.
   
   

   		2I		G =		9981		S ⇒ S =	G
   		1000				1000			499

   
   ​​Total resistance of Ammeter ​
   

   R =		=		G		G		=
   	SG			499					G
   	S + G			G		+ G			500
   				499

   Correct Option: C

   As 0.2% of main current passes through the galvanometer hence 998/1000 current through the shunt.
   
   

   		2I		G =		9981		S ⇒ S =	G
   		1000				1000			499

   
   ​​Total resistance of Ammeter ​
   

   R =		=		G		G		=
   	SG			499					G
   	S + G			G		+ G			500
   				499

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