Q 1 :

A 2 A current carrying straight metal wire of resistance 1 Ω, resistivity 2×10-6Ωm, area of cross-section 10 mm2 and mass 500 g is suspended horizontally in mid-air by applying a uniform magnetic field B. The magnitude of B is ____ ×10-1T (given, g = 10 m/s²)                           [2024]



(5)            R=ρlAl=RAρ=1×10×10-62×10-6=5m

                ilB=mgB=mgil

                B=(0.5)(10)2×5=5×10-1T

 



Q 2 :

A charge of 4.0 μC is moving with a velocity of 4.0×106ms-1 along the positive y-axis under a magnetic field B of strength (2k^) T. The force acting on the charge is xi^ N. The value of x is ____.                       [2024]



(32)            q=4μC,v=4×106j^m/s,    B=2k^T

                  F=q(v×B)=4×10-6(4×106j^×2k^)

                 F=4×10-6×8×106i^=32i^N

                 x=32

 



Q 3 :

A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immersed in a perpendicular magnetic field B=B0j^ as shown in the figure. The magnetic force on the wire if it has a current i is              [2024]

  • -2iBRj^

     

  • 2iBRj^

     

  • -iBRj^

     

  • iBRj^

     

(1)

Note: Direction of magnetic field is in +k^

Force on current carrying wire,

F=i×B

Here, =2R

F=-2iRBj^

 



Q 4 :

The magnetic field existing in a region is given by B=0.2(1+2x)k^T. A square loop of edge 50 cm carrying 0.5 A current is placed in the x – y plane with its edges parallel to the x – y axes, as shown in the figure. The magnitude of the net magnetic force experienced by the loop is ______ mN.                 [2024]



(50)

F=i(×B)

F3 and F4 will cancel each other.

F1=-0.5×50100×{0.2(1+4)}i^

=-12×12×1i^=-14i^=-0.25i^

F2=0.5×50100×{0.2(1+5)}i^=12×12×{1.2}i^=0.3i^

Fnet=0.05i^(N)=50 (mN)i^



Q 5 :

A square loop of edge length 2 m carrying a current of 2 A is placed with its edges parallel to the x-y axis. A magnetic field is passing through the x-y plane and expressed as B=B0(1+4x)k^,  where B0=5T.. The net magnetic force experienced by the loop is ______ N.              [2024]



(160)

B(x=0)=B0,  B(x=2)=9B0

Also, F=i×B

F1=iB0(j^×k^)=iB0(i^)=2×2×5(i^)=20N(i^)

F2=9iB0(-j^×k^)=9iB0(-i^)=2×2×45(-i^)=180N(-i^)

Fnet on AD and CB=0

Fnet=F2-F1=160N(-i^)

Fnet=160N



Q 6 :

The horizontal component of Earth's magnetic field at a place is 3.5×10-5T. A very long straight conductor carrying a current of 2A in the direction from South-East to North-West is placed. The force per unit length experienced by the conductor is ______ ×10-6N/m.         [2024]



(35)

F=IBsinθ

F=IBsinθ=2×3.5×10-5×sin45°

=2×3.5×10-5×12

=3.5×10-5=35×10-6 N/m

 



Q 7 :

A 4.0 cm long straight wire carrying a current of 8 A is placed perpendicular to an uniform magnetic field of strength 0.15 T. The magnetic force on the wire is ______ mN.          [2025]



(48)

F=IlB

F=8×4100×0.15=48×103 N= 48 mN



Q 8 :

A massless square loop of wire of resistance 10Ω supporting a mass of 1 g hangs vertically with one of its sides in a uniform magnetic field of 103 G, directed outwards in the shaded region. A dc voltage V is applied to the loop. For what value of V, the magnetic force will exactly balance the weight of the supporting mass of 1 g?          [2023]

(If sides of the loop = 10 cm, g=10 ms-2)

  • 110 V

     

  • 100 V

     

  • 1 V

     

  • 10 V

     

(4)

Fm=mg

  ILB=mg

  (vR)LB=mg

  v=mgRLB=(1×10-3 kg)(10 m/s2)(10Ω)(0.1 m)(103×10-4 T)=10 V



Q 9 :

Two long straight wires P and Q carrying equal current 10 A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F1. If the distance between the wires is halved and the currents on them are doubled, the force F2 on 10 cm length of wire P will be          [2023]

  • 10F1

     

  • 8F1

     

  • F18

     

  • F110

     

(2)

Force per unit length between two parallel straight wires

=μ0i1i22πd

F1F2=μ0(10)22π(5)μ0(20)22π(52)=18  F2=8F1



Q 10 :

A current carrying rectangular loop PQRS is made of uniform wire. The length PR=QS=5 cm and PQ=RS=100 cm. If the ammeter current reading changes from I to 2I, the ratio of magnetic forces per unit length on the wire PQ due to wire RS in the two cases respectively fPQI:fPQ2I is              [2023]

  • 1 : 2

     

  • 1 : 4

     

  • 1 : 5

     

  • 1 : 3

     

(2)

Force between two current-carrying wires=μ0I1I22πd×L

Here I1 and I2 are equal

F=μ0I22πd×L

FI2

FIF2I=I24I2=14



Q 11 :

An electron is moving along the positive x-axis. If the uniform magnetic field is applied parallel to the negative z-axis, then           [2023]

A. The electron will experience magnetic force along positive y-axis

B. The electron will experience magnetic force along negative y-axis

C. The electron will not experience any force in magnetic field

D. The electron will continue to move along the positive x-axis

E. The electron will move along circular path in magnetic field

Choose the correct answer from the options given below:

  • B and E only

     

  • A and E only

     

  • C and D only

     

  • B and D only

     

(1)

F=-e(v×B)

Force will be along the negative y-axis.

As magnetic force is perpendicular to velocity, the path of the electron must be a circle.



Q 12 :

A single turn current loop in the shape of a right angle triangle with sides 5 cm, 12 cm, 13 cm is carrying a current of 2 A. The loop is in a uniform magnetic field of magnitude 0.75 T whose direction is parallel to the current in the 13 cm side of the loop. The magnitude of the magnetic force x on the 5 cm side will be x130N. The value of x is _______.              [2023]



(9)

Force on 5 cm side is

|F|=ILBsinθ

       =(2)(5×10-2)×34×1213=9130N

So, x=9



Q 13 :

A straight wire AB of mass 40 g and length 50 cm is suspended by a pair of flexible leads in a uniform magnetic field of magnitude 0.40 T as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is _________ A.  (Take g=10 ms-2)                 [2023]



(2)

For equilibrium Mg=IlB

I=mgB=40×10-3×1050×10-2×0.4=2 A



Q 14 :

Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by 15 cm length of wire Q is _____. [2026]

  • 6×10-6 N towards R

     

  • 6×10-7 N towards R

     

  • 6×10-7 N towards P

     

  • 6×10-6 N towards P

     

(1)

Fnet=μ02πI0(I1d1+I2d2)

Fnet=2×10-7×(33+22)×15×10-210-2

=4×15×10-7

Fnet=6×10-6 N