JEE Physics Moving Charges and Magnetism MCQs Set 12

Answer: (a) \( \frac{\mu_0Q\omega}{4\pi r} \)

Explanation:
1. The rotating charge is equivalent to a current \( I=\frac{Q\omega}{2\pi} \).
2. Field at the centre of the ring: \( B=\frac{\mu_0I}{2r}=\frac{\mu_0}{2r}\cdot\frac{Q\omega}{2\pi} \).
3. \( B=\frac{\mu_0Q\omega}{4\pi r} \).

Teacher's Note:
1. A rotating charged ring is a circular current. First find the current from charge per unit time, then use the loop formula.

Answer: (c) \( \frac{\mu_0}{2\pi} \)

Explanation:
1. Right-hand rule: the field of the 5 A wire at P and the field of the 2.5 A wire at P are opposite in direction for currents in the same direction.
2. \( B_1=\frac{\mu_0(5)}{2\pi(2.5)}=\frac{\mu_0}{\pi} \) and \( B_2=\frac{\mu_0(2.5)}{2\pi(2.5)}=\frac{\mu_0}{2\pi} \).
3. \( B=B_1-B_2=\frac{\mu_0}{2\pi} \).

Teacher's Note:
1. Between two parallel wires carrying current in the same direction the fields oppose, so subtract. The larger current decides the direction.

Answer: (c) \( i_1\ell_1=i_2\ell_2 \)

Explanation:
1. The two arcs are in parallel across the same battery, so the potential difference across them is the same.
2. Resistance of a uniform wire is proportional to its length: \( R_1\propto\ell_1 \) and \( R_2\propto\ell_2 \).
3. \( i_1R_1=i_2R_2\Rightarrow i_1\ell_1=i_2\ell_2 \).

Teacher's Note:
1. Parallel branches share the same potential difference. Longer arcs have more resistance and carry less current.

Answer: (d) 90 T

Explanation:
1. For one turn of radius \( R \): \( B=\frac{\mu_0I}{2R}=10 \) T. The wire length is \( 2\pi R \).
2. For 3 turns the radius becomes \( R'=\frac R3 \).
3. \( B'=\frac{\mu_0(3)I}{2R'}=\frac{3\mu_0I}{2(R/3)}=9\times\frac{\mu_0I}{2R}=90 \) T.

Teacher's Note:
1. With the same wire, N turns means radius is 1/N times and field becomes N squared times larger. Do not forget both effects.

Answer: (c) independent of D

Explanation:
1. Total number of turns \( =nN \), so turns per unit length \( =\frac{nN}{L} \).
2. \( B=\mu_0\frac{nN}{L}i \).
3. It does not contain D, so the field is independent of D.

Teacher's Note:
1. In a long solenoid the field depends on the turns per unit length and the current only, not on the diameter.

Answer: (c) \( \vec E,\vec B \) and \( \vec v \) are mutually perpendicular and \( v=E/B \)

Explanation:
1. For no deflection \( q\vec E+q(\vec v\times\vec B)=0 \), so the electric and magnetic forces must be equal and opposite.
2. This needs \( \vec E,\vec B \) and \( \vec v \) to be mutually perpendicular with \( qE=qvB \).
3. Hence \( v=\frac EB \).

Teacher's Note:
1. This is a velocity selector. Only particles with speed E/B pass undeflected, whatever their charge or mass.

Answer: (b) 16 N in the \( -z \) direction

Explanation:
1. \( \vec F=q(\vec v\times\vec B) \) with \( \vec B=2\hat j \) T and \( q=-4\times10^{-6} \) C.
2. \( \vec v\times\vec B=(2\hat i+3\hat j)\times10^6\times2\hat j=4\times10^6\hat k \).
3. \( \vec F=-4\times10^{-6}\times4\times10^6\hat k=-16\hat k \) N, i.e. 16 N in the \( -z \) direction.

Teacher's Note:
1. Only the component of velocity perpendicular to B matters, and a negative charge reverses the direction of the force.

Answer: (a) 0.5 MeV

Explanation:
1. For a circular path \( r=\frac{\sqrt{2mK}}{qB} \). For the same \( r \) and \( B \): \( K\propto\frac{q^2}{m} \).
2. Proton: \( q=e,\ m=m_p \). Deuteron: \( q=e,\ m=2m_p \).
3. \( K_d=\frac{1}{2}K_p=0.5 \) MeV.

Teacher's Note:
1. Same charge and radius mean kinetic energy is inversely proportional to mass. The heavier deuteron needs less energy.

Answer: (d) \( \text{Li}^{++} \)

Explanation:
1. \( F=qvB\sin90^\circ=qvB \) for the same \( v \) and \( B \), so the force is proportional to the magnitude of the charge.
2. \( \text{Li}^{++} \) carries a charge of \( +2e \), while electron, proton and \( \text{He}^+ \) carry \( e \).
3. So Li++ experiences the maximum force.

Teacher's Note:
1. With the same speed and field the force depends only on the charge. Count the number of unit charges on each ion.

Answer: (c) \( 10^{-9} \) s

Explanation:
1. \( T=\frac{2\pi m}{eB}=\frac{2\pi}{(e/m)B} \).
2. \( T=\frac{2\pi}{1.76\times10^{11}\times3.57\times10^{-2}}=\frac{6.28}{6.28\times10^{9}} \).
3. \( T=10^{-9} \) s.

Teacher's Note:
1. The time period depends only on e/m and B, not on the speed or radius of the electron.

Answer: (c) \( \frac{\mu_0ne}{2r} \)

Explanation:
1. The current due to the revolving electron is \( I=ne \).
2. Field at the centre: \( B=\frac{\mu_0I}{2r} \).
3. \( B=\frac{\mu_0ne}{2r} \).

Teacher's Note:
1. A revolving charge is equivalent to a loop current equal to charge times frequency.

Answer: (a) zero

Explanation:
1. Both wires are 15 cm from the midpoint and carry equal currents in the same direction.
2. Their fields at the midpoint have equal magnitude \( \frac{\mu_0I}{2\pi(0.15)} \) but opposite directions.
3. The resultant is zero.

Teacher's Note:
1. Midway between two equal parallel currents in the same direction the fields cancel. For opposite currents the fields would add.

Answer: (b) \( \frac{Mg}{I\ell} \)

Explanation:
1. For equilibrium the upward magnetic force balances the weight: \( BI\ell=Mg \).
2. \( B=\frac{Mg}{I\ell} \).

Teacher's Note:
1. A horizontal wire in a horizontal perpendicular field can be levitated by a magnetic force equal to its weight.

Answer: (a) towards the left (towards P and Q)

Explanation:
1. Parallel currents attract. R is attracted towards Q and towards P, both on its left.
2. Force from Q on R per unit length: \( \frac{\mu_0(4)(6)}{2\pi r} \). Force from P: \( \frac{\mu_0(2)(6)}{2\pi(2r)} \). Both point to the left.
3. The resultant is towards the left.

Teacher's Note:
1. Currents in the same direction attract. When all other wires lie on one side, all forces point to that side.

Answer: (b) \( 4.5\times10^{-2} \) N m

Explanation:
1. Torque \( \tau=NIAB\sin\theta \), where \( \theta \) is the angle between the axis (magnetic moment) and the field.
2. \( \tau=1000\times6\times3\times10^{-4}\times5\times10^{-2}\times\sin30^\circ \).
3. \( 1000\times6\times3\times10^{-4}=1.8 \); \( 1.8\times5\times10^{-2}=0.09 \); \( 0.09\times0.5=4.5\times10^{-2} \) N m.

Teacher's Note:
1. The angle in the torque formula is the one between the dipole axis and the field, not between the plane and the field.

Answer: (a) can be in equilibrium in two orientations, one stable while the other is unstable

Explanation:
1. In a uniform field the torque \( \vec\tau=\vec M\times\vec B \) vanishes when \( \vec M \) is parallel or antiparallel to \( \vec B \).
2. Parallel (\( \theta=0 \)) is the minimum energy, stable position; antiparallel (\( \theta=180^\circ \)) is the unstable position.
3. So there are two equilibrium orientations, one stable and the other unstable.

Teacher's Note:
1. Equilibrium needs zero torque. Stable equilibrium is the minimum of potential energy, \( U=-MB\cos\theta \).

Answer: (c) \( 2\left(\frac{\tau}{\sqrt3Bi}\right)^{1/2} \)

Explanation:
1. Because \( \vec B \) lies in the plane of the loop, \( \sin\theta=1 \) and \( \tau=iAB \).
2. Area of the equilateral triangle \( A=\frac{\sqrt3}{4}\ell^2 \), so \( \tau=iB\frac{\sqrt3}{4}\ell^2 \).
3. \( \ell^2=\frac{4\tau}{\sqrt3Bi}\Rightarrow\ell=2\left(\frac{\tau}{\sqrt3Bi}\right)^{1/2} \).

Teacher's Note:
1. Torque on a loop is maximum when the field lies in the plane of the loop. Then \( \tau=iAB \).

Answer: (a) \( 3.0\times10^{-2} \) T

Explanation:
1. Mean radius \( r=\frac{25+26}{2}=25.5 \) cm \( =0.255 \) m.
2. Turns per unit length \( n=\frac{N}{2\pi r}=\frac{3500}{2\pi\times0.255} \).
3. \( B=\mu_0nI=4\pi\times10^{-7}\times\frac{3500}{2\pi\times0.255}\times11\approx3.0\times10^{-2} \) T.

Teacher's Note:
1. Use the mean radius to count the turns per unit length of a toroid. The field outside the toroid and in its hollow space is zero.

Answer: (b) 0.24 J

Explanation:
1. \( W=MB(\cos\theta_1-\cos\theta_2)=NIAB(\cos0^\circ-\cos180^\circ) \).
2. \( NIA=100\times0.1\times3.14\times(0.05)^2=0.0785 \) A m².
3. \( W=0.0785\times1.5\times2=0.2355\approx0.24 \) J.

Teacher's Note:
1. Work done equals the change in potential energy of the dipole, \( U=-MB\cos\theta \). A half turn from aligned to opposed changes U by 2MB.

Answer: (a) \( 3\times10^{-5} \) N/m, directed downward

Explanation:
1. The current (east to west) is perpendicular to the field (south to north), so \( F=I\ell B \) gives \( \frac Fl=IB=1\times3\times10^{-5} \) N/m.
2. Direction: \( \vec\ell\times\vec B \) with \( \vec\ell \) pointing west and \( \vec B \) pointing north is downward (west \( \times \) north = down).
3. So the force is \( 3\times10^{-5} \) N/m, downward.

Teacher's Note:
1. Use the vector product to get the direction. A current parallel to the field would feel no force at all.

Answer: (a) \( \frac{\mu_0nI}{4}(\sqrt3+1) \)

Explanation:
1. Axial field of a finite solenoid: \( B=\frac{\mu_0nI}{2}(\cos\theta_1-\cos\theta_2) \).
2. In the standard formula the angles are \( \theta_1=30^\circ \) and \( \theta_2=180^\circ-60^\circ=120^\circ \), so \( \cos30^\circ-\cos120^\circ=\frac{\sqrt3}{2}+\frac12 \).
3. \( B=\frac{\mu_0nI}{2}\cdot\frac{\sqrt3+1}{2}=\frac{\mu_0nI}{4}(\sqrt3+1) \).

Teacher's Note:
1. The formula contains the cosines of the angles measured from the axis. A negative cosine for an obtuse angle increases the field.