1. An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude:​​

1. 1. Zero ​

   
   
   

   2. μ0n²e

      r

   
   
   

   3. μ0ne

      2r

   
   
   

   4. μ0ne

      2πr

   
   
   

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1. View Hint View Answer Discuss in Forum

   Radius of circular orbit = r ​
   No. of rotations per second = n
   ​i.e., T = 1/n
   
   ​Magnetic field at its centre, B_c =? ​
   As we know, current ​
   

   i =	e	=	e	= en
   	T		(l/n)

   
   = equivalent current ​Magnetic field at the centre of circular orbit, ​

   Correct Option: C

   Radius of circular orbit = r ​
   No. of rotations per second = n
   ​i.e., T = 1/n
   
   ​Magnetic field at its centre, B_c =? ​
   As we know, current ​
   

   i =	e	=	e	= en
   	T		(l/n)

   
   = equivalent current ​Magnetic field at the centre of circular orbit, ​

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2. A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes

1. 1. ​inclined at 45° to the magnetic field ​

   
   
   

   2. ​inclined at any arbitrary angle to the magnetic field ​

   
   
   

   3. ​parallel to the magnetic field

   
   
   

   4. perpendicular to the magnetic field

   
   
   

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1. View Hint View Answer Discuss in Forum

   The plane of coil will orient itself so that area vector aligns itself along the magnetic field.So, the plane will orient perpendicular to the magnetic field.

   Correct Option: D

   The plane of coil will orient itself so that area vector aligns itself along the magnetic field.So, the plane will orient perpendicular to the magnetic field.

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3. A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B', at radial distances a/2 and 2a respectively, from the axis of the wire is :​​

1. 1. 1/4

   
   
   

   2. 1/2

   
   
   

   3. 1

   
   
   

   4. 4

   
   
   

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1. View Hint View Answer Discuss in Forum

   For points inside the wire i.e., (r ≤ R) ​
   

   Magnetic field B =	μ0Ir
   	2πR²

   
   For points outside the wire (r ≥ R)
   

   Magnetic field, B’ =	μ0I
   	2πR

   
   ​

   ∴		=	μ0I(a/2)		= 1 : 1
   	B		2πa²
   	B'		μ0I
   			2π(2a)

   Correct Option: C

   For points inside the wire i.e., (r ≤ R) ​
   

   Magnetic field B =	μ0Ir
   	2πR²

   
   For points outside the wire (r ≥ R)
   

   Magnetic field, B’ =	μ0I
   	2πR

   
   ​

   ∴		=	μ0I(a/2)		= 1 : 1
   	B		2πa²
   	B'		μ0I
   			2π(2a)

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4. A uniform magnetic field acts at right angles to the direction of motion of electron. As a result, the electron moves in a circular path of radius 2cm. If the speed of electron is doubled, then the radius of the circular path will be

1. 1. 2.0 cm

   
   
   

   2. 0.5 cm

   
   
   

   3. ​4.0 cm

   
   
   

   4. 1.0 cm

   
   
   

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1. View Hint View Answer Discuss in Forum

   r =	mv	or r ∝ v
   	qB

   
   As v is doubled, the radius also becomes double. Hence, radius = 2 × 2 = 4 cm

   Correct Option: C

   r =	mv	or r ∝ v
   	qB

   
   As v is doubled, the radius also becomes double. Hence, radius = 2 × 2 = 4 cm

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5. A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is :​​
   

1. 1. →B =	μ0		I	(μî × 2̂k)
      	4π		r

   
   
   

   2. →B = -	μ0		I	(μî + 2̂k)
      	4π		r

   
   
   

   3. →B =	μ0		I	(μî - 2̂k)
      	4π		r

   
   
   

   4. →B =	μ0		I	(μî + 2̂k)
      	4π		r

   
   
   

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1. View Hint View Answer Discuss in Forum

   Magnetic field due to segment ‘1’ ​
   

   →B1 =	μ0I	[sin 90° + sin 0°](-̂k)
   	4πR

   
   

   =	- μ0I	(-̂k) = →B3
   	4πR

   
   ​= ​Magnetic field due to segment 2 ​
   

   B2 =	μ0I	(-î) =	- μ0I	(πî)
   	4R		4πR

   
   
   ​∴ ^→_B at centre ​
   

   →Bc = →B1 + →B2 + →B3 =	- μ0I	((πî) + ̂k)
   	4πR

   
   

   Correct Option: B

   Magnetic field due to segment ‘1’ ​
   

   →B1 =	μ0I	[sin 90° + sin 0°](-̂k)
   	4πR

   
   

   =	- μ0I	(-̂k) = →B3
   	4πR

   
   ​= ​Magnetic field due to segment 2 ​
   

   B2 =	μ0I	(-î) =	- μ0I	(πî)
   	4R		4πR

   
   
   ​∴ ^→_B at centre ​
   

   →Bc = →B1 + →B2 + →B3 =	- μ0I	((πî) + ̂k)
   	4πR

   
   

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