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Magnetic Effect of Current

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

Q91
A horizontal overhead powerline is at height of 4m4m from the ground and carries a current of 100A100A from east to west. The magnetic field directly below it on the ground is (μ0=4π×10−7  Tm  A−1)\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,\,Tm\,\,{A^{ - 1}}} \right)
A 2.5×10−7 T2.5 \times {10^{ - 7}}\,T southward
B 5×10−6 T5 \times {10^{ - 6}}\,T northward
C 5×10−6 T5 \times {10^{ - 6}}\,T southward
D 2.5×10−7 T2.5 \times {10^{ - 7}}\,T northward
Correct Answer
Option C
Solution

The magnetic field is

B=μ04π2IrB = {{{\mu _0}} \over {4\pi }}{{2I} \over r}
=10−7×2×1004= {10^{ - 7}} \times {{2 \times 100} \over 4}
=5×10−6T= 5 \times {10^{ - 6}}T

According to right hand palm rule, the magnetic field is directed towards south.

Q92
A long solenoid is formed by winding 70 turns cm−1^{-1}. If 2.0 A current flows, then the magnetic field produced inside the solenoid is ____________ (μ0=4π×10−7\mu_0=4\pi\times10^{-7} TmA−1^{-1})
A 88×10−488\times10^{-4} T
B 1232×10−41232\times10^{-4} T
C 176×10−4176\times10^{-4} T
D 352×10−4352\times10^{-4} T
Correct Answer
Option C
Solution

Number of turns per meter =7000=7000 turns per m\mathrm{m}

i=2 AB=μ0ni=4π×10−7×7000×2=56π×10−4 T=56×227×10−4 T=176×10−4 T\begin{aligned} &i=2 \mathrm{~A} & \\\\ & B=\mu_{0} n i =4 \pi \times 10^{-7} \times 7000 \times 2 \\\\ & =56 \pi \times 10^{-4} \mathrm{~T} \\\\ & =56 \times \frac{22}{7} \times 10^{-4} \mathrm{~T} \\\\ & =176 \times 10^{-4} \mathrm{~T} \end{aligned}
Q93
The magnetic field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point along the axis at a distance r2{r \over 2} from the centre is :
A B/2
B 2B
C (25)3B{\left( {{2 \over {\sqrt 5 }}} \right)^3}B
D (23)3B{\left( {{2 \over {\sqrt 3}}} \right)^3}B
Correct Answer
Option C
Solution
B=μ0I2rB = {{{\mu _0}I} \over {2r}}
Ba=μ0Ir22(r2+r24){B_a} = {{{\mu _0}I{r^2}} \over {2\left( {{r^2} + {{{r^2}} \over 4}} \right)}}
⇒BaB=(25)3\Rightarrow {{{B_a}} \over B} = {\left( {{2 \over {\sqrt 5 }}} \right)^3}
⇒Ba=(25)3B\Rightarrow {B_a} = {\left( {{2 \over {\sqrt 5 }}} \right)^3}B
Q94
Two identical conducting wires AOBAOB and CODCOD are placed at right angles to each other. The wire AOBAOB carries an electric current I1{I_1} and CODCOD carries a current I2{I_2}. The magnetic field on a point lying at a distance dd from OO, in a direction perpendicular to the plane of the wires AOBAOB and CODCOD , will be given by
A μ02πd(I12+I22){{{\mu _0}} \over {2\pi d}}\left( {I_1^2 + I_2^2} \right)
B μ02π(I1+I2d)12{{{\mu _0}} \over {2\pi }}{\left( {{{{I_1} + {I_2}} \over d}} \right)^{{1 \over 2}}}
C μ02πd(I12+I22)12{{{\mu _0}} \over {2\pi d}}{\left( {I_1^2 + I_2^2} \right)^{{1 \over 2}}}
D μ02πd(I1+I2){{{\mu _0}} \over {2\pi d}}\left( {{I_1} + {I_2}} \right)
Correct Answer
Option C
Solution

Clearly, the magnetic fields at a point

P,P,

equidistant from

AOBAOB

and

CODCOD

will have directions perpendicular to each other, as they are placed normal to each other. ∴\therefore Resultant field,

B=B12+B22B = \sqrt {B_1^2 + B_2^2}

But

B1=μ0I12πd{B_1} = {{{\mu _0}{I_1}} \over {2\pi d}}

and

B2=μ0I22πd{B_2} = {{{\mu _0}{I_2}} \over {2\pi d}}

∴\therefore

B=(μ02πd)2(I12+I22)B = \sqrt {{{\left( {{{{\mu _0}} \over {2\pi d}}} \right)}^2}\left( {I_1^2 + I_2^2} \right)}

or,

B=μ02πd(I12+I22)1/2B = {{{\mu _0}} \over {2\pi d}}{\left( {I_1^2 + I_2^2} \right)^{1/2}}
Q95
A circular loop of radius rr is carrying current I A. The ratio of magnetic field at the center of circular loop and at a distance r from the center of the loop on its axis is :
A 32\sqrt2 : 2
B 1 : 32\sqrt2
C 22\sqrt2 : 1
D 1 : 2\sqrt2
Correct Answer
Option C
Solution
BP1=μ0l2rBP2=μ0lr22(r2+r2)3/2=μ0I25/2r\begin{aligned} B_{P_{1}} & =\frac{\mu_{0} l}{2 r} \\\\ B_{P_{2}} & =\frac{\mu_{0} l r^{2}}{2\left(r^{2}+r^{2}\right)^{3 / 2}}=\frac{\mu_{0} I}{2^{5 / 2} r} \end{aligned}
∴BP1BP2=μ0I2rμ0I25/2r=221\begin{aligned} & \therefore \quad \frac{B_{P_{1}}}{B_{P_{2}}}=\frac{\frac{\mu_{0} I}{2 r}}{\frac{\mu_{0} I}{2^{5 / 2} r}}=\frac{2 \sqrt{2}}{1} \end{aligned}
Q96
A beam of protons with speed 4 × 105 ms–1 enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to : (Mass of the proton = 1.67 × \times 10–27 kg, charge of the proton = 1.69 × \times 10–19 C)
A 2 cm
B 12 cm
C 5 cm
D 4 cm
Correct Answer
Option D
Solution

Pitch =

2πmqB\frac{2\pi m}{qB}

vcosθ\theta =

2(3.14)(1.67×10−27)×4×105×cos⁡60(1.69×10−19)(0.3){{2\left( {3.14} \right)\left( {1.67 \times {{10}^{ - 27}}} \right) \times 4 \times {{10}^5} \times \cos 60} \over {\left( {1.69 \times {{10}^{ - 19}}} \right)\left( {0.3} \right)}}

= 0.04 m = 4 cm

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