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Magnetic Effect of Current

NEET Physics · 96 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11
A very long conducting wire is bent in a semi-circular shape from AA to BB as shown in figure. The magnetic field at point PP for steady current configuration is given by :
A μ0i4R\dfrac{\mu_{0} \mathrm{i}}{4 \mathrm{R}} pointed away from the page
B μ0i4R[1−2π]\dfrac{\mu_{0} \mathrm{i}}{4 \mathrm{R}}\left[1-\dfrac{2}{\pi}\right] pointed away from page
C μ0i4R[1−2π]\dfrac{\mu_{0} \mathrm{i}}{4 \mathrm{R}}\left[1-\dfrac{2}{\pi}\right] pointed into the page
D μ0i4R\dfrac{\mu_{0} \mathrm{i}}{4 \mathrm{R}} pointed into the page
Correct Answer
Option B
Solution
B=μ04πIR(π)−μ04π2IR\mathrm{B=\frac{\mu_{0}}{4 \pi} \frac{I}{R}(\pi)-\frac{\mu_{0}}{4 \pi} \frac{2 I}{R}}
=μ0I4R[1−2π]=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{R}}\left[1-\frac{2}{\pi}\right]

outward i.e. away from page.

Q12
A closely packed coil having 1000 turns has an average radius of 62.8 cm. If current carried by the wire of the coil is 1 A, the value of magnetic field produced at the centre of the coil will be (permeability of free space =4π×10−7 = 4\pi \times {10^{ - 7}} H/m) nearly
A 10 −-3 T
B 10 −-1 T
C 10 −-2 T
D 10 2 T
Correct Answer
Option A
Solution

Magnetic field at the centre of coil

B=μ0NI2RB = {{{\mu _0}NI} \over {2R}}

On substituting the given values

⇒B=4π×10−7×1032×62.8×10−2\Rightarrow B = {{4\pi \times {{10}^{ - 7}} \times {{10}^3}} \over {2 \times 62.8 \times {{10}^{ - 2}}}}
=10−3 T= {10^{ - 3}}\,T
Q13
The shape of the magnetic field lines due to an infinite long, straight current carrying conductor is
A a plane
B a straight line
C circular
D elliptical
Correct Answer
Option C
Solution

From the right hand curl rule, the shape of magnetic field lines due to long current carrying wire is circular (concentric circles)

Q14
Two very long, straight, parallel conductors A and B carry current of 5 A and 10 A respectively and are at a distance of 10 cm from each other. The direction of current in two conductors is same. The force acting per unit length between two conductors is : (μ\mu 0 = 4π\pi ×\times 10 −-7 SI unit)
A 1 ×\times 10 −-4 Nm −-1 and is repulsive
B 2 ×\times 10 −-4 Nm −-1 and is attractive
C 2 ×\times 10 −-4 Nm −-1 and is repulsive
D 1 ×\times 10 −-4 Nm −-1 and is attractive
Correct Answer
Option D
Solution

Force per unit length between two long current carrying wires

=μ0l1l22πd= {{{\mu _0}{l_1}{l_2}} \over {2\pi d}}
F=4π×10−7×5×102π×10×10−2=10−4F = {{4\pi \times {{10}^{ - 7}} \times 5 \times 10} \over {2\pi \times 10 \times {{10}^{ - 2}}}} = {10^{ - 4}}

Nm −-1 ⇒\Rightarrow The force is attractive as direction of current in two conductors is same Hence, required force is

1×10−41 \times {10^{ - 4}}

Nm −-1 and it is attractive.

Q15
The magnetic field on the axis of a circular loop of radius 100 cm carrying current I=2 AI = \sqrt 2 \,A, at point 1 m away from the centre of the loop is given by :
A 6.28×10−46.28 \times {10^{ - 4}} T
B 3.14×10−73.14 \times {10^{ - 7}} T
C 6.28×10−76.28 \times {10^{ - 7}} T
D 3.14×10−43.14 \times {10^{ - 4}} T
Correct Answer
Option B
Solution

Magnetic field at the axis of a current carrying circular loop

∣B∣=μ0ia22[a2+d2]3/2⇒∣B∣=4π×10−7×2×12[1+1]3/2|B| = {{{\mu _0}i{a^2}} \over {2{{[{a^2} + {d^2}]}^{3/2}}}} \Rightarrow |B| = {{4\pi \times {{10}^{ - 7}} \times \sqrt 2 \times 1} \over {2{{[1 + 1]}^{3/2}}}}
∣B∣=4π×10−7×22.(2)3/2=4π×10−7(2)3/2×(2)1/2=3.14×10−7|B| = {{4\pi \times {{10}^{ - 7}} \times \sqrt 2 } \over {2.{{(2)}^{3/2}}}} = {{4\pi \times {{10}^{ - 7}}} \over {{{(2)}^{3/2}} \times {{(2)}^{1/2}}}} = 3.14 \times {10^{ - 7}}

T

Q16
A long solenoid of radius 1 mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is
A 6.28 ×\times 10 −-2 T
B 12.56 ×\times 10 −-2 T
C 12.56 ×\times 10 −-4 T
D 6.28 ×\times 10 −-4 T
Correct Answer
Option B
Solution

We know, magnetic field at centre of solenoid

B=μ0Nll=μ0nlB = {\mu _0}{N \over l}l = {\mu _0}nl
[n=Nl]\left[ {n = {N \over l}} \right]
=4π×10−7×100×103×1= 4\pi \times {10^{ - 7}} \times 100 \times {10^3} \times 1
[n=10010−3]\left[ {n = {{100} \over {{{10}^{ - 3}}}}} \right]
=4π×10−2T= 4\pi \times {10^{ - 2}}T
B=12.56×10−2TB = 12.56 \times {10^{ - 2}}T
Q17
From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is
A Uniform and remains constant for both the regions.
B A linearly increasing function of distance upto the boundary of the wire and then linearly decreasing for the outside region.
C A linearly increasing function of distance r upto the boundary of the wire and then decreasing one with 1r{1 \over r} dependence for the outside region.
D A linearly decreasing function of distance upto the boundary of the wire and then a linearly increasing one of the outside region.
Correct Answer
Option C
Solution

For solid wire Inside point

B=μ0Ir2R2×2πrB = {{{\mu _0}I{r^2}} \over {{R^2} \times 2\pi {r}}}
=μ0IrR2×2π= {{{\mu _0}Ir} \over {{R^2} \times 2\pi }}
B∝rB \propto r

Outside point

B=μ0I2πrB = {{{\mu _0}I} \over {2\pi r}}
B∝1rB \propto {1 \over r}
Q18
An infinitely long straight conductor carries a current of 5A as shown. An electron is moving with a speed of 10 5 m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.
A 8 ×\times 10 −-20 N
B 4 ×\times 10 −-20 N
C 8π\pi ×\times 10 −-20 N
D 4π\pi ×\times 10 −-20 N
Correct Answer
Option A
Solution
B=μ0I2πRB = {{{\mu _0}I} \over {2\pi R}}

F = BVq sinθ\theta θ\theta = 90

∘^\circ

F = BVq

F=μ0I2πR×V×e=2×10−7×520×10−2×105×1.6×10−19F = {{{\mu _0}I} \over {2\pi R}} \times V \times e = {{2 \times {{10}^{ - 7}} \times 5} \over {20 \times {{10}^{ - 2}}}} \times {10^5} \times 1.6 \times {10^{ - 19}}

F = 8 ×\times 10 −-20 N

Q19
A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is : (μ0=4π×10−7TmA−1)\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}Tm{{A}^{-1}}} \right)
A 3.14 ×10−4T{ \times {{10}^{ - 4}}T}
B 6.28 ×10−5T{ \times {{10}^{ - 5}}T}
C 3.14 ×10−5T{ \times {{10}^{ - 5}}T}
D 6.28 ×10−4T{ \times {{10}^{ - 4}}T}
Correct Answer
Option D
Solution

Magnetic field at centre of solenoid =

μ0nI{\mu _0}nI
n=NL=10050×10−2n = {N \over L} = {{100} \over {50 \times {{10}^{ - 2}}}}

= 200 turns/m I = 25A Now, substituting the values, B = 4π\pi ×\times 10 -7 ×\times 200 ×\times 2.5 = 6.28 ×\times 10 -4 T

Q20
Ionized hydrogen atoms and α\alpha -particles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths r H : r α\alpha will be :
A 1 : 2
B 4 : 1
C 1 : 4
D 2 : 1
Correct Answer
Option D
Solution

Radius of the path (r) =

mvqB{{mv} \over {qB}}

=

pqB{p \over {qB}}

So, the radius for H-atom r H =

peB{p \over {eB}}

The radius for α\alpha particle r α\alpha =

p2eB{p \over {2eB}}

∴\therefore

rHrα=peBp2eB{{{r_H}} \over {{r_\alpha }}} = {{{p \over {eB}}} \over {{p \over {2eB}}}}

=

21{2 \over 1}
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