Question 101:
easyA uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron
Since the velocity \(vec{v}\) is parallel to the magnetic field \(vec{B}\), the magnetic force is zero. The electric field exerts a force opposite to the direction of velocity on the negatively charged electron, decreasing its speed.
Question 102:
easy\(\varepsilon_0\) and \(\mu_0\) are the electric permittivity and magnetic permeability of free space respectively. If the corresponding quantities of a medium are \(2\varepsilon_0\) and \(1.5\mu_0\) respectively, the refractive index of the medium will nearly be
The speed of light in free space is \(c = \frac{1}{\sqrt{\varepsilon_0\mu_0}}\) and in a medium is \(v = \frac{1}{\sqrt{\varepsilon\mu}}\). The refractive index \(n = \frac{c}{v} = \sqrt{\frac{\varepsilon\mu}{\varepsilon_0\mu_0}} = \sqrt{2 \times 1.5} = \sqrt{3}\).
Question 103:
easyA long straight wire of length 2 m and mass 250 g is suspended horizontally in a uniform horizontal magnetic field of 0.7 T. The amount of current flowing through the wire will be (\(g = 9.8\text{ ms}^{-2}\))
For the wire to be suspended, the upward magnetic force must equal gravity: \(I L B = mg ⇒ I = \frac{mg}{LB}\). Plugging in the values: \(I = \frac{0.250 \times 9.8}{2 \times 0.7} = 1.75\text{ A}\).
Question 104:
easyFor the hollow thin cylindrical current carrying straight pipe which statement is correct:
Inside the hollow pipe, the magnetic field is zero according to Ampere's law. Since a current-carrying pipe is electrically neutral, the electric field outside the pipe is zero.
Question 105:
moderateA helium nucleus is moving in a circular path of radius \(0.8\text{ m}\). If it takes \(2\text{ sec}\) to complete one revolution, the magnetic field produced at the centre of the circle is:
Current is \(I = \frac{q}{T} = \frac{2e}{2} = e = 1.6 \times 10^{-19}\text{ A}\). The magnetic field at the centre is \(B = \frac{\mu_0 I}{2R} = \frac{\mu_0 (1.6 \times 10^{-19})}{2(0.8)} = \mu_0 \times 10^{-19}\text{ T}\).
Question 106:
easyWhich of the following charges has the maximum frequency of revolution in a uniform transverse magnetic field?
Frequency of revolution in a magnetic field is given by \(f = \frac{qB}{2\pi m}\). Since the electron has the highest charge-to-mass ratio \(q/m\) among the charged particles, it has the maximum frequency.
Question 107:
easyA thin circular wire carrying a current \(I\) has a magnetic moment \(M\). The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment of:
For circle: \(M = I \pi R^2\) where \(2\pi R = L ⇒ R = \frac{L}{2\pi}\), so \(M = \frac{I L^2}{4\pi}\). For square of side \(a = \frac{L}{4}\): \(M' = I a^2 = \frac{I L^2}{16}\). Thus, \(M' = \frac{\pi}{4} M\).
Question 108:
easyA particle of mass \(M\) and charge \(Q\) moving with velocity \(\vec{v}\) describes a circular path of radius \(R\) when subjected to a uniform transverse magnetic field of induction \(B\). The work done by the field when the particle completes one full circle is:
The magnetic force \(\vec{F} = Q(\vec{v} \times \vec{B})\) is always perpendicular to the velocity \(\vec{v}\). Therefore, power \(P = \vec{F} \cdot \vec{v} = 0\), meaning the work done is always zero.
Question 109:
easyA magnetic field:
The magnetic force is given by \(F = qvB\sin\theta\). If the particle moves across the field lines, \(\sin\theta \neq 0\), resulting in a non-zero force.
Question 110:
moderateIf a charged particle goes unaccelerated in a region containing electric & magnetic fields:
(a) \(\vec{E}\) must be perpendicular to \(\vec{B}\)
(b) \(\vec{v}\) must be perpendicular to \(\vec{E}\)
(c) \(\vec{v}\) must be perpendicular to \(\vec{B}\)
(d) \(E\) must be equal to \(vB\)
For the net force to be zero, \(\vec{F}_e + \vec{F}_b = 0 ⇒ \vec{E} = -(\vec{v} \times \vec{B})\). Since \(\vec{E}\) is the cross product, it must be perpendicular to both \(\vec{B}\) and \(\vec{v}\).