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Magnetic Effect of Current

JEE Physics · 124 questions · Page 1 of 13 · Click an option or "Show Solution" to reveal answer

Q1
In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a
A helix
B straight line
C ellipse
D circle
Correct Answer
Option B
Solution

The charged particle will move along the lines of electric field (and magnetic field).

Magnetic field will exert no force.

The force by electric field will be along the lines of uniform electric field.

Hence the particle will move in a straight line.

Q2
A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicualr to the plane. Let rp, re and rHe be their respective radii, then
A re < rp < rHe
B re < rp = rHe
C re > rp > rHe
D re > rp = rHe
Correct Answer
Option B
Solution
r=mvqB=2mKqBr = {{mv} \over {qB}} = {{\sqrt {2mK} } \over {qB}}
rHe=rp>re{r_{He}} = {r_p} > {r_e}
Q3
A rectangular coil (Dimension 5 cm × 2.5 cm) with 100 turns, carrying a current of 3 A in the clock-wise direction is kept centered at the origin and in the X-Z plane. A magnetic field of 1 T is applied along X-axis. If the coil is tilted through 45° about Z-axis, then the torque on the coil is :
A 0.42 Nm
B 0.55 Nm
C 0.38 Nm
D 0.27 Nm
Correct Answer
Option D
Solution
∣τ→∣=∣M‾×B‾∣\left| {\overrightarrow \tau } \right| = \left| {\overline M \times \overline B } \right|
τ=NI×A×B×sin⁡45o\tau = NI \times A \times B \times \sin {45^o}
τ=0.27 Nm\tau = 0.27 \,Nm
Q4
Two ions having same mass have charges in the ratio 1 : 2. They are projected normally in a uniform magnetic field with their speeds in the ratio 2 : 3. The ratio of the radii of their circular trajectories is :
A 1 : 4
B 4 : 3
C 3 : 1
D 2 : 3
Correct Answer
Option B
Solution
R=mvqB⇒R1R2=mv1q1Bmv2q2B=v1q1×q2v2=q2q1×v1v2R = {{mv} \over {qB}} \Rightarrow {{{R_1}} \over {{R_2}}} = {{{{m{v_1}} \over {{q_1}B}}} \over {{{m{v_2}} \over {{q_2}B}}}} = {{{v_1}} \over {{q_1}}} \times {{{q_2}} \over {{v_2}}} = {{{q_2}} \over {{q_1}}} \times {{{v_1}} \over {{v_2}}}
=21×(23)=43= {2 \over 1} \times \left( {{2 \over 3}} \right) = {4 \over 3}
Q5
A charge particle moves along circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charge particle increases to 4 times its initial value. What will be the ratio of new radius to the original radius of circular path of the charge particle :
A 1 : 1
B 1 : 2
C 2 : 1
D 1 : 4
Correct Answer
Option C
Solution
R=mvBq=2mKBqR = {{mv} \over {Bq}} = {{\sqrt {2mK} } \over {Bq}}
⇒R∝K\Rightarrow R \propto \sqrt K

⇒\Rightarrow ratio = 2 : 1

Q6
The electric current in a circular coil of four turns produces a magnetic induction 32 T at its centre. The coil is unwound and is rewound into a circular coil of single turn, the magnetic induction at the centre of the coil by the same current will be :
A 2 T
B 4 T
C 8 T
D 16 T
Correct Answer
Option A
Solution

By given information

32=4×μ0i2r32 = 4 \times {{{\mu _0}i} \over {2r}}

..... (i) Also,

r′=4rr' = 4r

...... (ii) and

B′=1×μ0i2r′B' = 1 \times {{{\mu _0}i} \over {2r'}}

.... (iii)

⇒B′=μ0i2(4r)=μ0i8r=18×16=2 T\Rightarrow B' = {{{\mu _0}i} \over {2(4r)}} = {{{\mu _0}i} \over {8r}} = {1 \over 8} \times 16 = 2\,T
Q7
Consider a long thin conducting wire carrying a uniform current I. A particle having mass "M" and charge " qq " is released at a distance " aa " from the wire with a speed v0v_0 along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance xx from the wire. The value of xx is [ μ0\mu_0 is vacuum permeability]
A a[1−mvo2qμ0I]a\left[1-\dfrac{m v_o}{2 q \mu_0 \mathrm{I}}\right]
B ae−4πmvoqμoIa e^{-\dfrac{4 \pi \mathrm{mv}_o}{q \mu_o \mathrm{I}}}
C a[1−mvqμoI]a\left[1-\dfrac{\mathrm{mv}}{\mathrm{q}} \mu_{\mathrm{o}} \mathrm{I}\right]
D a2\dfrac{a}{2}
Correct Answer
Option B
Solution

For

A→BA \to B
V→=−vxi^+vyj^\overrightarrow V = - {v_x}\widehat i + {v_y}\widehat j
B→=μ0I2πr(−k^)\overrightarrow B = {{{\mu _0}I} \over {2\pi r}}\left( { - \widehat k} \right)

∵\because

F→=q(v→×B→)\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)

So,

F→=qμ0I2πr[[(−vx)i^+vyj^]×(−k^)]\overrightarrow F = {{q{\mu _0}I} \over {2\pi r}}\left[ {\left[ {\left( { - {v_x}} \right)\widehat i + {v_y}\widehat j} \right] \times \left( { - \widehat k} \right)} \right]
=qμ0I2πr[vx(i^×k^)−vy(j^×k^)]= {{q{\mu _0}I} \over {2\pi r}}\left[ {{v_x}\left( {\widehat i \times \widehat k} \right) - {v_y}\left( {\widehat j \times \widehat k} \right)} \right]
F→=qμ0I2πr[−vxj^−vyi^]\overrightarrow F = {{q{\mu _0}I} \over {2\pi r}}\left[ { - {v_x}\widehat j - {v_y}\widehat i} \right]
⇒Fx=−μ0Iq2πrvy\Rightarrow {F_x} = {{ - {\mu _0}Iq} \over {2\pi r}}{v_y}
⇒max=−μ0Iq2πrvy\Rightarrow m{a_x} = {{ - {\mu _0}Iq} \over {2\pi r}}{v_y}
⇒vxdvxdr=−μ0Iq2πmvyr\Rightarrow {v_x}{{d{v_x}} \over {dr}} = {{ - {\mu _0}Iq} \over {2\pi m}}{{{v_y}} \over r}
⇒vxdvxvy=−μ0Iq2πmdrr\Rightarrow {{{v_x}d{v_x}} \over {{v_y}}} = {{ - {\mu _0}Iq} \over {2\pi m}}{{dr} \over r}

Since,

vx2+vy2=v02v_x^2 + v_y^2 = v_0^2
⇒vy=v02−vx2\Rightarrow {v_y} = \sqrt {v_0^2 - v_x^2}
⇒∫0v0vxdvxv02−vx2=−μ0Iq2πm∫ax′drr\Rightarrow \int_0^{{v_0}} {{{{v_x}d{v_x}} \over {\sqrt {v_0^2 - v_x^2} }} = {{ - {\mu _0}Iq} \over {2\pi m}}\int_a^{x'} {{{dr} \over r}} }

Let

v02−vx2=z2v_0^2 - v_x^2 = {z^2}
⇒−2vxdvx=2zdz\Rightarrow - 2{v_x}d{v_x} = 2zdz

when

vx=0, z=v0{v_x} = 0,\,z = {v_0}
vx=v0, z=0{v_x} = {v_0},\,z = 0
⇒∫v00zdzz=−μ0Iq2πm∫ax′drr\Rightarrow \int_{{v_0}}^0 {{{zdz} \over z} = - {{{\mu _0}Iq} \over {2\pi m}}\int_a^{x'} {{{dr} \over r}} }
⇒−v0=μ0Iq2πmln⁡(x′a)\Rightarrow - {v_0} = {{{\mu _0}Iq} \over {2\pi m}}\ln \left( {{{x'} \over a}} \right)
⇒ln⁡(x′a)=−2πmv0μ0Iq\Rightarrow \ln \left( {{{x'} \over a}} \right) = - {{2\pi m{v_0}} \over {{\mu _0}Iq}}
⇒x′=ac−2πmv0μ0Iq\Rightarrow x' = ac{{ - 2\pi m{v_0}} \over {{\mu _0}Iq}}

.... (1) Now, for

B→CB \to C
v→=−vxi^−vyj^\overrightarrow v = - {v_x}\widehat i - {v_y}\widehat j
B→=μ0I2πr(−k^)\overrightarrow B = {{{\mu _0}I} \over {2\pi r}}\left( { - \widehat k} \right)
F→=q(v→×B→)=qμ0I2πr[vx(i^×k^)+vy(j^×k^)]\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right) = {{q{\mu _0}I} \over {2\pi r}}\left[ {{v_x}\left( {\widehat i \times \widehat k} \right) + {v_y}\left( {\widehat j \times \widehat k} \right)} \right]
⇒F→=μ0Iq2πr[−vxj^+vyi^]\Rightarrow \overrightarrow F = {{{\mu _0}Iq} \over {2\pi r}}\left[ { - {v_x}\widehat j + {v_y}\widehat i} \right]
⇒Fx=μ0Iq2πrvy\Rightarrow {F_x} = {{{\mu _0}Iq} \over {2\pi r}}{v_y}
⇒vxdvxdr=μ0Iq2πmvyr\Rightarrow {v_x}{{d{v_x}} \over {dr}} = {{{\mu _0}Iq} \over {2\pi m}}{{{v_y}} \over r}
∫v00vxdvxv02−vx2=μ0Iq2πm∫x′xdrr\int_{{v_0}}^0 {{{{v_x}d{v_x}} \over {\sqrt {v_0^2 - v_x^2} }} = {{{\mu _0}Iq} \over {2\pi m}}\int_{x'}^x {{{dr} \over r}} }

Let

v02−vx2=z2v_0^2 - v_x^2 = {z^2}
−2vxdvx=2zdz- 2{v_x}d{v_x} = 2zdz
vx=v0⇒z=0{v_x} = {v_0} \Rightarrow z = 0
vx=0⇒z=v0{v_x} = 0 \Rightarrow z = {v_0}
⇒−∫0v0zdzz=μ0Iq2πm[ln⁡r]x′x\Rightarrow - \int_0^{{v_0}} {{{zdz} \over z} = {{{\mu _0}Iq} \over {2\pi m}}\left[ {\ln r} \right]_{x'}^x}
⇒−v0=μ0Iq2πmln⁡(xx′)\Rightarrow - {v_0} = {{{\mu _0}Iq} \over {2\pi m}}\ln \left( {{x \over {x'}}} \right)
⇒x=x′e−2πmv0μ0Iq\Rightarrow x = x'{e^{ - {{2\pi m{v_0}} \over {{\mu _0}Iq}}}}

.... (2) From (1) & (2),

x=ae−4πmv0μ0Iqx = a{e^{ - {{4\pi m{v_0}} \over {{\mu _0}Iq}}}}
Q8
A solenoid of 1000 turns per metre has a core with relative permeability 500. Insulated windings of the solenoid carry an electric current of 5A. The magnetic flux density produced by the solenoid is : (permeability of free space = 4π\pi ×\times 10−-7 H/m)
A π\piT
B 2 ×\times 10−-3π\pi T
C 10−-4π\pi T
D π5{\pi \over 5} T
Correct Answer
Option A
Solution

B = μ\mu n i B = μ\mur μ\mu0 n i B = 500 ×\times 4π\pi ×\times 10−-7 ×\times 103 ×\times 5 B = π\pi ×\times 10−-3 ×\times 103 B = π\pi T

Q9
A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of magnetic field will be equal to
A B
B 2B
C 4B
D B2{B \over 2}
Correct Answer
Option A
Solution
B=μ0niB = {\mu _0}ni

Now

i→2ii \to 2i

And

n→n2n \to {n \over 2}
B′=μ0n2×2i=μ0ni=BB' = {\mu _0}{n \over 2} \times 2i = {\mu _0}ni = B
Q10
A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r (r
A B ∝\propto r2
B B ∝\propto r
C B ∝\propto 1r2{1 \over {{r^2}}}
D B ∝\propto 1r{1 \over {{r}}}
Correct Answer
Option B
Solution
∫B‾ . dl‾=μ0Iin\int {\overline B \,.\,\overline {dl} = {\mu _0}{I_{in}}}
⇒B×2πr=μ0IπR2×πr2\Rightarrow B \times 2\pi r = {{{\mu _0}I} \over {\pi {R^2}}} \times \pi {r^2}
⇒B∝r\Rightarrow B \propto r
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