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Magnetic Effect of Current

JEE Physics · 124 questions · Page 4 of 13 · Click an option or "Show Solution" to reveal answer

Q31
A current II flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius R.R. The magnitude of the magnetic induction along its axis is:
A μ0I2π2R{{{\mu _0}I} \over {2{\pi ^2}R}}
B μ0I2πR{{{\mu _0}I} \over {2\pi R}}
C μ0I4πR{{{\mu _0}I} \over {4\pi R}}
D μ0Iπ2R{{{\mu _0}I} \over {{\pi ^2}R}}
Correct Answer
Option D
Solution

Current in a small element,

dl=dθπIdl = {{d\theta } \over \pi }I

Magnetic field due to the element

dB=μ04π2dlRdB = {{{\mu _0}} \over {4\pi }}{{2dl} \over R}

The component

dBdB
cos⁡ θ,\cos \,\theta ,

of the field is canceled by another opposite component. Therefore,

Bnet=∫dBsin⁡θ=μ0I2π2R0{B_{net}} = \int {dB\sin \theta = {{{\mu _0}I} \over {2{\pi ^2}{R_0}}}}
∫0πsin⁡θdθ=μ0Iπ2R\int\limits_0^\pi {\sin \theta d\theta = {{{\mu _0}I} \over {{\pi ^2}R}}}
Q32
Two identical wires AA and B,B, each of length ′l′'l', carry the same current II. Wire AA is bent into a circle of radius RR and wire BB is bent to form a square of side ′a′'a'. If BA{B_A} and BB{B_B} are the values of magnetic fields at the centres of the circle and square respectively, then the ratio BABB{{{B_A}} \over {{B_B}}} is:
A π216{{{\pi ^2}} \over {16}}
B π282{{{\pi ^2}} \over {8\sqrt 2 }}
C π28{{{\pi ^2}} \over {8}}
D π2162{{{\pi ^2}} \over {16\sqrt 2 }}
Correct Answer
Option B
Solution

Case (a) :

BA=μ04πIR×2π{B_A} = {{{\mu _0}} \over {4\pi }}{I \over R} \times 2\pi
=μ04πIℓ/2π×2π= {{{\mu _0}} \over {4\pi }}{I \over {\ell /2\pi }} \times 2\pi
         \,\,\,\,\,\,\,\,\,
(2πR=ℓ)\left( {2\pi R = \ell } \right)
=μ04πIℓ×(2π)2= {{{\mu _0}} \over {4\pi }}{I \over \ell } \times {\left( {2\pi } \right)^2}

Case (b) :

BB=4×μ04πIa/2   {B_B} = 4 \times {{{\mu _0}} \over {4\pi }}{I \over {a/2}}\,\,\,
[sin⁡  45∘+sin⁡  45∘]\left[ {\sin \,\,{{45}^ \circ } + \sin \,\,{{45}^ \circ }} \right]
=4×μ04π×Iℓ/8×22= 4 \times {{{\mu _0}} \over {4\pi }} \times {I \over {\ell /8}} \times {2 \over {\sqrt 2 }}
=μ0I4π ℓ×232      [4a=1]= {{{\mu _0}I} \over {4\pi \,\ell }} \times \sqrt[{32}]2 \,\,\,\,\,\,\left[ {4a = 1} \right]
Q33
A negative test charge is moving near a long straight wire carrying a current. The force acting on the test charge is parallel to the direction of the current. The motion of the charge is :
A away from the wire
B towards the wire
C parallel to the wire along the current
D parallel to the wire opposite to the current
Correct Answer
Option B
Solution

Given situation is shown in the figure As we know,

F→=q(v→×B→)\overrightarrow F = q(\overrightarrow v \times \overrightarrow B )
F→=−q0(v→×B→)\overrightarrow F = - {q_0}(\overrightarrow v \times \overrightarrow B )

According to question, direction of current is parallel to the force acting on the electron.

Hence, the motion of test charge is towards the wire.

Q34
In a certain region static electric and magnetic fields exist. The magnetic field is given by B→=B0(i^+2j^−4k^)\overrightarrow B = {B_0}\left( {\widehat i + 2\widehat j - 4\widehat k} \right) . If a test charge moving with a velocity υ→=υ0(3i^−j^+2k^)\overrightarrow \upsilon = {\upsilon _0}\left( {3\widehat i - \widehat j + 2\widehat k} \right) experiences no force in that region, then the electric field in the region, in SI units, is :
A E→=−υ0 B0(3i^−2j^−4k^)\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {3\widehat i - 2\widehat j - 4\widehat k} \right)
B E→=−υ0 B0(i^+j^+7k^)\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {\widehat i + \widehat j + 7\widehat k} \right)
C E→=υ0 B0(14j^+7k^)\overrightarrow E = {\upsilon _0}\,{B_0}\left( {14\widehat j + 7\widehat k} \right)
D E→=−υ0 B0(14j^+7k^)\overrightarrow E = - {\upsilon _0}\,{B_0}\left( {14\widehat j + 7\widehat k} \right)
Correct Answer
Option D
Solution

Here test charge experience no net force, So, sum of electric and magnetic field is zero.

∴   \therefore\,\,\,

Fe + Fm = 0

∴   \therefore\,\,\,

Fe = −- q (

v→\overrightarrow v

×\times

B→)\overrightarrow B)

= −- qB0

υ\upsilon

0 [(3

i^\widehat i

−-

j^\widehat j

+ 2

k^\widehat k

) ×\times (

i^\widehat i

+ 2

j^\widehat j

−- 4

k^\widehat k

)] = −- q

υ\upsilon

0 B0 (14

j^\widehat j

+ 7

k^\widehat k

) Electric field produced by the charge q,

E→\overrightarrow E

=

Fe→q{{\overrightarrow {{F_e}} } \over q}

=

−qυ0B0(14j^+7k^)q{{ - q{\upsilon _0}{B_0}\left( {14\widehat j + 7\widehat k} \right)} \over q}

= −-

υ\upsilon

0 B0 (14

j^{\widehat j}

+ 7

k^{\widehat k}

)

Q35
A magnetic dipole in a constant magnetic field has :
A maximum potential energy when the torque is maximum.
B zero potential energy when the torque is minimum.
C zero potential energy when the torque is maximum.
D minimum potential energy when the torque is maximum.
Correct Answer
Option C
Solution

In uniform magnetic field, the torque experienced by the magnetic dipole is τ\tau = MB sin θ\theta Torque will be maximum when θ\theta = 90o τ\taumax = MB sin90o = MB Potential energy of magnetic dipole, μ\mu = −- MB cos θ\theta at maximum torque, μ\mu = −- MB cos 90o = 0

Q36
A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity ω\omega with resect to normal axis then the magnetic moment of the loop is :
A q ω\omega r2
B 43{4 \over 3} q ω\omega r2
C 32{3 \over 2} q ω\omega r2
D 12{1 \over 2} q ω\omega r2
Correct Answer
Option D
Solution

Magnetic moment, μ\mu = I A =

qT(πr2){q \over T}\left( {\pi {r^2}} \right)

=

q2π/ω(πr2){q \over {2\pi /\omega }}\left( {\pi {r^2}} \right)

=

qw2π{{qw} \over {2\pi }}
(πr2)\left( {\pi {r^2}} \right)

=

12{1 \over 2}

qω\omegar2

Q37
A current of 1 A is flowing on the sides of an equilateral triangle of side 4.5 × \times 10-2 m. The magnetic field at the center of the triangle will be :
A 2 × \times 10-5 Wb/m2
B Zero
C 8 × \times 10-5 Wb/m2
D 4 × \times 10-5 Wb/m2
Correct Answer
Option D
Solution

We know that magnetic field due to finite current carrying wire is

B=μ04πIb(cos⁡θ1+cos⁡θ2)B = {{{\mu _0}} \over {4\pi }}{I \over b}(\cos {\theta _1} + \cos {\theta _2})

.....

(1) Given that side of triangle = 4.5 ×\times 10−-2 m = a; current = 1A since the triangle is equilateral, angle of each side will be 60

∘^\circ

. Now,

tan⁡θ=PerpendicularBase⇒tan⁡60∘=a/2b\tan \theta = {{Perpendicular} \over {Base}} \Rightarrow \tan 60^\circ = {{a/2} \over b}
⇒b=a2tan⁡60∘=a23\Rightarrow b = {a \over {2\tan 60^\circ }} = {a \over {2\sqrt 3 }}

Using equation (1), we get

B=μ04πIa/23(cos⁡30∘+cos⁡30∘)=μ04π23Ia2cos⁡30∘B = {{{\mu _0}} \over {4\pi }}{I \over {a/2\sqrt 3 }}(\cos 30^\circ + \cos 30^\circ ) = {{{\mu _0}} \over {4\pi }}{{2\sqrt 3 I} \over a}2\cos 30^\circ
B=μ04π23Ia2×32=μ04π6IaB = {{{\mu _0}} \over {4\pi }}{{2\sqrt 3 I} \over a}{{2 \times \sqrt 3 } \over 2} = {{{\mu _0}} \over {4\pi }}{{6I} \over a}
⇒B=10−7×6×14.5×10−2=1.33×10−5∼2×10−5\Rightarrow B = {{{{10}^{ - 7}} \times 6 \times 1} \over {4.5 \times {{10}^{ - 2}}}} = 1.33 \times {10^{ - 5}} \sim 2 \times {10^{ - 5}}

Wb/m2

Q38
An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii re, rp, rα_\alpha respectively in a uniform magnetic field B. The relation between re, rp, rα_\alpha is:
A re < rα_\alpha < rp
B re > rp = rα_\alpha
C re < rp = rα_\alpha
D re < rp < rα_\alpha
Correct Answer
Option C
Solution

When a charged particle moves in a magnetic field then the charged particle moves in a circular path. So,

mv2r{{m{v^2}} \over r}

= Bqv ⇒\Rightarrow

   \,\,\,

r =

mvBq{{mv} \over {Bq}}

We know kinetic energy, K =

12{1 \over 2}

mv2

∴   \therefore\,\,\,

mv =

2Km\sqrt {2Km}
∴   \therefore\,\,\,

r =

2KmBq{{\sqrt {2Km} } \over {Bq}}

According to the question, Ke (electron) = Kp (proton) = Kα\alpha(Alpha particle) = K = constant, and all of them are in uniform magnetic field.

∴\therefore B = constant.

∴   \therefore\,\,\,

r ∝\propto

mq{{\sqrt m } \over q}

For proton (1H1), mass = m, and charge = e

∴   \therefore\,\,\,

rp ∝\propto

me{{\sqrt m } \over e}

For alpha particle (2H4), mass = 4m and charge = 2e

∴   \therefore\,\,\,

rα\alpha ∝\propto

4m2e{{\sqrt {4m} } \over {2e}}

∝\propto

me{{\sqrt m } \over e}

∴\therefore

   \,\,\,

rp = rα\alpha For electron, charge = e and mass (me) = 9.1 ×\times 10−-31 kg and mass of proton = 1.67 ×\times 10−-27 kg

∴   \therefore\,\,\,

mass of electron < mass of proton. re ∝\propto

mee{{\sqrt {{m_e}} } \over e}

< rp

∴   \therefore\,\,\,

re < rp = r∝\propto

Q39
A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. if this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:
A mπ{m \over \pi }
B 3mπ{{3m} \over \pi }
C 2mπ{{2m} \over \pi }
D 4mπ{{4m} \over \pi }
Correct Answer
Option D
Solution

Let the given square loop has side aa, then its magnetic dipole moment will be

m=Ia2m=I a^2

When square is converted into a circular loop of radius rr, Then, wire length will be same in both area,

⇒4a=2πr⇒r=4a2π=2aπ\Rightarrow 4 a=2 \pi r \Rightarrow r=\frac{4 a}{2 \pi}=\frac{2 a}{\pi}

Hence, area of circular loop formed is, A′=πr2A^{\prime}=\pi r^2

=π(2aπ)2=4a2π=\pi\left(\frac{2 a}{\pi}\right)^2=\frac{4 a^2}{\pi}

Magnitude of magnetic dipole moment of circular loop will be

m′=IA′=I4a2πm^{\prime}=I A^{\prime}=I \frac{4 a^2}{\pi}

Ratio of magnetic dipole moments of both shapes is,

m′m=I⋅4a2πIa2=4π⇒m′=4mπ(A−m)\begin{aligned} \frac{m^{\prime}}{m}=\frac{I \cdot \frac{4 a^2}{\pi}}{I a^2}=\frac{4}{\pi} \\\\ \Rightarrow m^{\prime} =\frac{4 m}{\pi}(\mathrm{A}-\mathrm{m}) \end{aligned}
Q40
A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40 π\pi rad s–1 about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8 × 10–9 T, then the charge carried by the ring is close to (μ\mu 0 = 4π\pi × 10–7 N/A2 ).
A 7 × 10–6 C
B 4 × 10–5 C
C 2 × 10–6 C
D 3 × 10–5 C
Correct Answer
Option D
Solution
B=μ0i2aωq2π=iB = {{{\mu _0}i} \over {2a}}{{\omega q} \over {2\pi }} = i
B=μ02a.ωq2πB = {{{\mu _0}} \over {2a}}.{{\omega q} \over {2\pi }}
B=10−7×400.1×q×πB = {{{{10}^{ - 7}} \times 40} \over {0.1}} \times q \times \pi
⇒q=3×10−5C\Rightarrow q = 3 \times {10^{ - 5}}C
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