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51.

Assertion : A beam of protons is moving towards east in vertically upward magnetic field. Then, this beam will deflect towards south. Reason : A constant magnetic force will act on the proton beam.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true.

Answer» Correct Answer - C
By changing the directioin of velocity direction of magnetic fore will change. So it is not a constant force.
Discussion
52.

A stream of protons and `alpha`-particle of equal momenta enter a unifom magnetic field perpendicularly. The radii of their orbits are in the ratioA. `1:1`B. `1:2`C. `2:1`D. `4:1`

Answer» Correct Answer - C
`r=P/(Bq) or rprop1/q` (as P=constant)`
Discussion
53.

As loop of magnetic moment `M` is placed in the orientation of unstable equilbirum position in a uniform magnetic field `B`. The external work done in rotating it through an angle `theta` isA. `-MB(1-costheta)`B. `-Mbcostheta`C. `Mbcostheta`D. `MB(1-costheta)`

Answer» Correct Answer - A
`theta_i=180^@, theta_f=180^@-theta`
`W=U_f-U_i`
`=-Mbcos(180^@-theta)-(-MBcos180^@)`
`=Mbcostheta-MB`
Discussion
54.

A square loop `OABCO` of side of side `l` carries a current `i`. It is placed as shown in figure. Find the magnetic moment of loop.

Answer» As discussed above, magnetic moment of the loop can be written as
`M=i(BCxxCO)`
`Here, BC=-lhatk`, `CO=-cos60^[email protected]^@hatj=l/2hati-sqrt(3l)/2hatj`
`:. M=i[(-lhatk)xx(-l/2hati-sqrt(3l)/2hatj)]`
or `M=(il^2)/2(hatj-sqrt3hati)`
Discussion
55.

The magnetic dipole moment of current loop is independent ofA. number of turnsB. area of loopC. current in the loopD. masgnetic field in which it is lying

Answer» Correct Answer - D
`M=NiA`
Discussion
56.

A current `l` flows along the length of an infinitely long, straight, thin-walled pipe. Then, (a) the magnetic field at all points inside the pipe is the same, but not zero (b) the magnetic field at any point inside the pipe is zero (c) the magnetic field is zero only on the axis of the pipe (d) the magnetic field is different at different points inside the pipe

Answer» Using Amperels circuital law over a circular loop of any radius less than the radius of the pipe, we can see that net curent inside the loop is zero. Hence, magnetic field at every point inside the loop will be zero.
Discussion
57.

Four very long, current carrying wires in the same plane intersect to form a square `40.0 cm` on each side as shown in figure. Find the magnitude and direction of the current `I` so that the magnetic field at the centre of square is zero. Wires are insulated from each other.

Answer» Correct Answer - A::B::D
`10A` and `8A` current produce inward magnetic field. While `20 A` current produces outward magnetic field. Hence, current in fourth wire should be `(20-10-8)A`. or `2A` and it should produce inward magnetic field. So, it should be downwards tward the bottom.
Discussion
58.

Two long straight wires, each carrying a current `I` in opossite directions asre separated by a distasnce `R`. The magnetic induction at a point mid way between the wire isA. zeroB. `(mu_0I)/(piR)`C. `(2mu_I)/(piR)`D. `(mu_0I)/(4piR)`

Answer» Correct Answer - C
Two fields are additive.
`:. B_("net")=2[(mu_0I)/(2pi(R/2)]=(2mu_0I)/(piR)]`
Discussion
59.

A very long wire carrying a current `I = 5.0 A` is bent at right angles. Find the magnetic induction at a point lying on a perpendicular normal to the plane of the wire drawn through the Point of bending at a distance `l = 35 cm` from it.

Answer» Correct Answer - B
`B_(net)=sqrt(B^2+B^2)=sqrt2B`
where `B=mu_0/(4pi) i/r(sin0^@+sin90^@)`
Discussion
60.

The figure shows tow coaxial circulalr loop a1 and 2, which forms same solid angle `theta` at point `O`. If `B_1` and `B_2` are the magnetic fields produced at the point `O` due to loop 1 and 2 respectively, then A. `B_1/B_2=1`B. `B_1/B_2=2`C. `B_1/B_2=8`D. `B_1/B_2=4`

Answer» Correct Answer - B
`B_1 =(muIR^3)/(2(R^2+x^2)^(3/2))`
`B_2=(mu_I(2R)^2)/(2[(2R)^2+(2x)^2]^(3/2))`
`B_1/B_2=2`
Discussion
61.

The figure shows three indentical current carryng square loops `A, B` and `C`. Identify the correct statement related to magnetic field B at the centre `O` of the square loop. Current in each wire is I. A. B is zero in all casesB. B is zero only incase f CC. B is non zero in all casesD. B is non zero only in case of B

Answer» Correct Answer - B
In c two, wires are producing `(o.)` magnitude field and two wires are producing `(ox)` magnetic field.
Discussion
62.

The straight wire `AB` carries a current `I`. The ends of the wire subtend angles `theta_1` and `theta_2` at the point `P` as shown in figure. The magnetic field at the point `P` is A. `(mu0I)/(4pia)(sintheta_1-sintheta_2)`B. `(mu_0I)/(4pia)(sintheta_1+sintheta_2)`C. `(mu_0I)/(4pia)(costheta_1-costheta_2)`D. `(mu_0I)/(4pia)(costheta_1-costheta_2)`

Answer» Correct Answer - A
NA
Discussion
63.

A wire along the x-axis carries a current of `3.50A` in the negative direction. Calculate the force (expressed in term of unit vectors) on a `1.00cm` sectioin of the wire exerted by these magnetic fiels. a. `B=-(0.65T)hatj b. `B=+(0.56T)hatk c. `B=-(0.31T)hatj` d. `B=_(0.33T)hati-(0.28t)hatk` e. `B=+(0.74T)hatj-0(0.36T)hatk`

Answer» `because i=3.5a`
`I=(-10^-2hatj)`
Now, apply `F=i(IxxB)` in all parts.
Discussion
64.

What is the value of `B` that can be set up at the equator to permit a proton of speed `10^7m//s` to circulate around the earth? `[R=6.4xx10^6m,m_p=1.67xx10^-27kg]`

Answer» From the relation
`r=(mv)/(Bq)`
We have
`B=(mv)/(qr)`
Substituting the vaues we have
`B=((1.67xx10^-27)(10^7))/((1.6xx10^-19)(6.4xx10^6))`
`=1.6xx10^-8T`
Discussion